CH3 – Avancement d\u2019une transformation chimique<\/title>\n<style>\n:root{--ink:#1f2937;--blue:#1d4e89;--red:#9b1c31;--gold:#d9a441;--soft:#f6f8fb;--line:#d7e0ea;--green:#2f7d59;}\n*{box-sizing:border-box} body{margin:0;font-family:Inter,system-ui,Segoe UI,Arial,sans-serif;color:var(--ink);background:linear-gradient(180deg,#f7fafc,#fff);line-height:1.55}\nheader{background:linear-gradient(120deg,#15324f,#1d4e89);color:white;padding:34px 24px;text-align:center} h1{margin:0;font-size:clamp(1.8rem,4vw,3rem)} header p{opacity:.92;margin:.5rem auto;max-width:850px}\nnav{position:sticky;top:0;background:white;border-bottom:1px solid var(--line);z-index:10;padding:10px;display:flex;gap:8px;flex-wrap:wrap;justify-content:center} nav a,.btn{padding:9px 12px;border-radius:999px;text-decoration:none;background:#eef4fb;color:#123b63;font-weight:700;border:1px solid #d7e4f4} .btn.red{background:#fff1f2;color:#8b1e3f}.btn.green{background:#ecfdf3;color:#166534}\nmain{max-width:1100px;margin:24px auto;padding:0 18px}.card{background:white;border:1px solid var(--line);box-shadow:0 10px 30px rgba(20,40,60,.08);border-radius:18px;padding:22px;margin:18px 0}h2{color:var(--blue);border-left:6px solid var(--gold);padding-left:12px;margin-top:0}h3{color:var(--red)}.important{background:#fff8e6;border-left:5px solid var(--gold);padding:12px 14px;border-radius:12px}.method{background:#f1f7ff;border-left:5px solid var(--blue);padding:12px 14px;border-radius:12px}.formula{font-weight:800;text-align:center;background:#f7f9fc;border:1px dashed #9fb3c8;border-radius:12px;padding:12px;margin:12px 0;font-size:1.12rem}.grid{display:grid;grid-template-columns:repeat(auto-fit,minmax(260px,1fr));gap:14px}table{width:100%;border-collapse:collapse;margin:12px 0;background:white}th,td{border:1px solid #c9d5e3;padding:8px;text-align:center}th{background:#eaf2fb;color:#123b63}.small{font-size:.92rem}.pill{display:inline-block;padding:2px 8px;border-radius:999px;background:#eef4fb;color:#123b63;font-weight:700}.quiz details{border:1px solid var(--line);border-radius:12px;margin:8px 0;padding:10px;background:#fff}summary{cursor:pointer;font-weight:800;color:#123b63}.svgbox{background:#fbfdff;border:1px solid var(--line);border-radius:14px;padding:10px;overflow:auto}.audio{display:flex;gap:8px;flex-wrap:wrap;align-items:center}.warn{background:#fff1f2;border-left:5px solid var(--red);padding:12px;border-radius:12px}\n@media print{nav,.audio,.buttons{display:none}.card{box-shadow:none;break-inside:avoid}body{background:white}}\n<\/style>\n<\/head><body>\n<header><h1>CH3 – Avancement d\u2019une transformation chimique<\/h1><p>Tableau d\u2019avancement, avancement maximal, r\u00e9actif limitant et composition finale. Version enrichie \u00e0 partir de la trace \u00e9crite.<\/p><\/header>\n<nav><a href=\"#cours\">Cours<\/a><a href=\"#methode\">M\u00e9thode<\/a><a href=\"#exos\">Exercices int\u00e9gr\u00e9s<\/a><a href=\"#bac\">Applications bac<\/a><a href=\"#quiz\">Quiz<\/a><a class=\"btn red\" href=\"https:\/\/pcwallis.malo.wf\/wp-content\/uploads\/2026\/06\/chapitre_avancement_premiere_cours.pdf\">PDF cours<\/a><a class=\"btn green\" href=\"https:\/\/pcwallis.malo.wf\/wp-content\/uploads\/2026\/06\/chapitre_avancement_premiere_exercices_types.pdf\">PDF exercices<\/a><\/nav>\n<main>\n<section class=\"card audio\"><strong>Lecture audio :<\/strong><button onclick=\"readPage()\">Lire le cours<\/button><button onclick=\"speechSynthesis.cancel()\">Stop<\/button><span class=\"small\">Utilise la synth\u00e8se vocale du navigateur.<\/span><\/section>\n<section id=\"cours\" class=\"card\"><h2>1. R\u00e9action chimique<\/h2><p>Une r\u00e9action chimique se produit entre des esp\u00e8ces chimiques. Les <strong>r\u00e9actifs<\/strong> sont consomm\u00e9s et les <strong>produits<\/strong> sont form\u00e9s.<\/p><div class=\"formula\">a A + b B \u2192 c C + d D<\/div><p>A et B sont les r\u00e9actifs ; C et D sont les produits ; a, b, c et d sont les coefficients st\u0153chiom\u00e9triques. Une \u00e9quation correctement ajust\u00e9e conserve les \u00e9l\u00e9ments chimiques et la charge.<\/p><div class=\"important\"><strong>Exemple de ta trace \u00e9crite :<\/strong> CH<sub>4<\/sub> + 2 O<sub>2<\/sub> \u2192 CO<sub>2<\/sub> + 2 H<sub>2<\/sub>O.<\/div><\/section>\n<section class=\"card\" id=\"methode\"><h2>2. Avancement et tableau d\u2019avancement<\/h2><p>L\u2019avancement <strong>x<\/strong> est une grandeur en mol qui permet de suivre l\u2019\u00e9volution des quantit\u00e9s de mati\u00e8re pendant une transformation.<\/p><table><tr><th>\u00c9tat<\/th><th>CH<sub>4<\/sub><\/th><th>O<sub>2<\/sub><\/th><th>CO<sub>2<\/sub><\/th><th>H<sub>2<\/sub>O<\/th><\/tr><tr><td>Initial<\/td><td>n<sub>0<\/sub>(CH<sub>4<\/sub>)<\/td><td>n<sub>0<\/sub>(O<sub>2<\/sub>)<\/td><td>0<\/td><td>0<\/td><\/tr><tr><td>En cours<\/td><td>n<sub>0<\/sub>(CH<sub>4<\/sub>) \u2212 x<\/td><td>n<sub>0<\/sub>(O<sub>2<\/sub>) \u2212 2x<\/td><td>x<\/td><td>2x<\/td><\/tr><tr><td>Final<\/td><td>n<sub>0<\/sub>(CH<sub>4<\/sub>) \u2212 x<sub>max<\/sub><\/td><td>n<sub>0<\/sub>(O<sub>2<\/sub>) \u2212 2x<sub>max<\/sub><\/td><td>x<sub>max<\/sub><\/td><td>2x<sub>max<\/sub><\/td><\/tr><\/table><div class=\"method\"><strong>Ta m\u00e9thode \u00e0 conserver :<\/strong> on calcule n<sub>0<\/sub>\/\u03bd pour chaque r\u00e9actif. Le plus petit r\u00e9sultat donne x<sub>max<\/sub> et le r\u00e9actif limitant.<\/div><div class=\"formula\">x<sub>max<\/sub> = min( n<sub>0<\/sub>(A)\/a ; n<sub>0<\/sub>(B)\/b )<\/div><\/section>\n<section class=\"card\"><h2>3. Sch\u00e9ma de raisonnement<\/h2><div class=\"svgbox\"><svg width=\"820\" height=\"230\" viewBox=\"0 0 820 230\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><defs><marker id=\"arrow\" markerWidth=\"10\" markerHeight=\"10\" refX=\"8\" refY=\"3\" orient=\"auto\"><path d=\"M0,0 L0,6 L9,3 z\" fill=\"#1d4e89\"\/><\/marker><\/defs><rect x=\"20\" y=\"55\" width=\"145\" height=\"80\" rx=\"15\" fill=\"#eaf2fb\" stroke=\"#1d4e89\"\/><text x=\"92\" y=\"88\" text-anchor=\"middle\" font-weight=\"700\">\u00c9quation<\/text><text x=\"92\" y=\"112\" text-anchor=\"middle\">ajust\u00e9e<\/text><line x1=\"165\" y1=\"95\" x2=\"235\" y2=\"95\" stroke=\"#1d4e89\" stroke-width=\"3\" marker-end=\"url(#arrow)\"\/><rect x=\"235\" y=\"55\" width=\"145\" height=\"80\" rx=\"15\" fill=\"#fff8e6\" stroke=\"#d9a441\"\/><text x=\"308\" y=\"88\" text-anchor=\"middle\" font-weight=\"700\">Quantit\u00e9s<\/text><text x=\"308\" y=\"112\" text-anchor=\"middle\">initiales<\/text><line x1=\"380\" y1=\"95\" x2=\"450\" y2=\"95\" stroke=\"#1d4e89\" stroke-width=\"3\" marker-end=\"url(#arrow)\"\/><rect x=\"450\" y=\"55\" width=\"145\" height=\"80\" rx=\"15\" fill=\"#f1f7ff\" stroke=\"#1d4e89\"\/><text x=\"523\" y=\"88\" text-anchor=\"middle\" font-weight=\"700\">Tableau<\/text><text x=\"523\" y=\"112\" text-anchor=\"middle\">d\u2019avancement<\/text><line x1=\"595\" y1=\"95\" x2=\"665\" y2=\"95\" stroke=\"#1d4e89\" stroke-width=\"3\" marker-end=\"url(#arrow)\"\/><rect x=\"665\" y=\"55\" width=\"135\" height=\"80\" rx=\"15\" fill=\"#fff1f2\" stroke=\"#9b1c31\"\/><text x=\"733\" y=\"86\" text-anchor=\"middle\" font-weight=\"700\">x<tspan baseline-shift=\"sub\">max<\/tspan><\/text><text x=\"733\" y=\"112\" text-anchor=\"middle\">\u00e9tat final<\/text><text x=\"410\" y=\"180\" text-anchor=\"middle\" fill=\"#9b1c31\" font-weight=\"700\">On ne choisit jamais le r\u00e9actif limitant au hasard : on compare n\u2080 \/ coefficient.<\/text><\/svg><\/div><\/section>\n<section id=\"exos\" class=\"card\"><h2>4. Exercice int\u00e9gr\u00e9 type cours<\/h2><p>On fait r\u00e9agir 3,0 \u00d7 10<sup>\u22122<\/sup> mol de CH<sub>4<\/sub> avec 5,0 \u00d7 10<sup>\u22122<\/sup> mol de O<sub>2<\/sub>.<\/p><details><summary>Voir la correction guid\u00e9e<\/summary><p>Pour CH<sub>4<\/sub> : x = 3,0 \u00d7 10<sup>\u22122<\/sup> \/ 1 = 3,0 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><p>Pour O<sub>2<\/sub> : x = 5,0 \u00d7 10<sup>\u22122<\/sup> \/ 2 = 2,5 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><p>Donc O<sub>2<\/sub> est le r\u00e9actif limitant et x<sub>max<\/sub> = 2,5 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><p>\u00c9tat final : CH<sub>4<\/sub> restant = 5,0 \u00d7 10<sup>\u22123<\/sup> mol ; O<sub>2<\/sub> = 0 ; CO<sub>2<\/sub> = 2,5 \u00d7 10<sup>\u22122<\/sup> mol ; H<sub>2<\/sub>O = 5,0 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><\/details><\/section>\n<section id=\"bac\" class=\"card\"><h2>5. Projection bac : r\u00e9action redox et couleur finale<\/h2><p>On exploite une r\u00e9action donn\u00e9e et on applique exactement la m\u00eame m\u00e9thode d\u2019avancement.<\/p><div class=\"formula\">MnO<sub>4<\/sub><sup>\u2212<\/sup> + 5 Fe<sup>2+<\/sup> + 8 H<sup>+<\/sup> \u2192 Mn<sup>2+<\/sup> + 5 Fe<sup>3+<\/sup> + 4 H<sub>2<\/sub>O<\/div><p>Les ions MnO<sub>4<\/sub><sup>\u2212<\/sup> sont violets. Si MnO<sub>4<\/sub><sup>\u2212<\/sup> reste \u00e0 la fin, la solution reste violette. S\u2019il est totalement consomm\u00e9, la couleur violette dispara\u00eet.<\/p><div class=\"warn\"><strong>Comp\u00e9tence bac :<\/strong> d\u00e9terminer les quantit\u00e9s initiales \u00e0 partir de C \u00d7 V, construire le tableau, identifier le r\u00e9actif limitant, puis interpr\u00e9ter la couleur finale.<\/div><\/section>\n<section id=\"quiz\" class=\"card quiz\"><h2>6. Quiz de fin<\/h2><details><summary>1. Que repr\u00e9sente x ?<\/summary><p>L\u2019avancement de la r\u00e9action, exprim\u00e9 en mol.<\/p><\/details><details><summary>2. Comment trouve-t-on le r\u00e9actif limitant ?<\/summary><p>On compare n<sub>0<\/sub>\/\u03bd pour chaque r\u00e9actif. Le plus petit r\u00e9sultat correspond au r\u00e9actif limitant.<\/p><\/details><details><summary>3. Pourquoi diviser par le coefficient st\u0153chiom\u00e9trique ?<\/summary><p>Parce que la r\u00e9action consomme les r\u00e9actifs dans les proportions donn\u00e9es par l\u2019\u00e9quation ajust\u00e9e.<\/p><\/details><details><summary>4. Que permet la ligne finale du tableau ?<\/summary><p>Elle donne la composition finale du syst\u00e8me : r\u00e9actifs restants et produits form\u00e9s.<\/p><\/details><details><summary>5. Dans la r\u00e9action avec MnO<sub>4<\/sub><sup>\u2212<\/sup>, que signifie une couleur violette finale ?<\/summary><p>Il reste des ions permanganate MnO<sub>4<\/sub><sup>\u2212<\/sup> : ils n\u2019\u00e9taient pas limitants.<\/p><\/details><\/section>\n<\/main><script>function readPage(){speechSynthesis.cancel();const t=document.querySelector('main').innerText;const u=new SpeechSynthesisUtterance(t);u.lang='fr-FR';u.rate=.95;speechSynthesis.speak(u);}<\/script><\/body><\/html>\n","protected":false},"excerpt":{"rendered":"<p>CH3 – Avancement d\u2019une transformation chimique CH3 – Avancement d\u2019une transformation chimique Tableau d\u2019avancement, avancement maximal, r\u00e9actif limitant et composition finale. Version enrichie \u00e0 partir de...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-961","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/961","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=961"}],"version-history":[{"count":5,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/961\/revisions"}],"predecessor-version":[{"id":971,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/961\/revisions\/971"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=961"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":961,"date":"2026-06-02T11:19:50","date_gmt":"2026-06-02T09:19:50","guid":{"rendered":"https:\/\/pcwallis.malo.wf\/?page_id=961"},"modified":"2026-06-02T12:41:49","modified_gmt":"2026-06-02T10:41:49","slug":"tableau-davancement","status":"publish","type":"page","link":"https:\/\/pcwallis.malo.wf\/index.php\/tableau-davancement\/","title":{"rendered":"tableau d’avancement"},"content":{"rendered":"\n\n\n\n\n\nCH3 – Avancement d\u2019une transformation chimique<\/title>\n<style>\n:root{--ink:#1f2937;--blue:#1d4e89;--red:#9b1c31;--gold:#d9a441;--soft:#f6f8fb;--line:#d7e0ea;--green:#2f7d59;}\n*{box-sizing:border-box} body{margin:0;font-family:Inter,system-ui,Segoe UI,Arial,sans-serif;color:var(--ink);background:linear-gradient(180deg,#f7fafc,#fff);line-height:1.55}\nheader{background:linear-gradient(120deg,#15324f,#1d4e89);color:white;padding:34px 24px;text-align:center} h1{margin:0;font-size:clamp(1.8rem,4vw,3rem)} header p{opacity:.92;margin:.5rem auto;max-width:850px}\nnav{position:sticky;top:0;background:white;border-bottom:1px solid var(--line);z-index:10;padding:10px;display:flex;gap:8px;flex-wrap:wrap;justify-content:center} nav a,.btn{padding:9px 12px;border-radius:999px;text-decoration:none;background:#eef4fb;color:#123b63;font-weight:700;border:1px solid #d7e4f4} .btn.red{background:#fff1f2;color:#8b1e3f}.btn.green{background:#ecfdf3;color:#166534}\nmain{max-width:1100px;margin:24px auto;padding:0 18px}.card{background:white;border:1px solid var(--line);box-shadow:0 10px 30px rgba(20,40,60,.08);border-radius:18px;padding:22px;margin:18px 0}h2{color:var(--blue);border-left:6px solid var(--gold);padding-left:12px;margin-top:0}h3{color:var(--red)}.important{background:#fff8e6;border-left:5px solid var(--gold);padding:12px 14px;border-radius:12px}.method{background:#f1f7ff;border-left:5px solid var(--blue);padding:12px 14px;border-radius:12px}.formula{font-weight:800;text-align:center;background:#f7f9fc;border:1px dashed #9fb3c8;border-radius:12px;padding:12px;margin:12px 0;font-size:1.12rem}.grid{display:grid;grid-template-columns:repeat(auto-fit,minmax(260px,1fr));gap:14px}table{width:100%;border-collapse:collapse;margin:12px 0;background:white}th,td{border:1px solid #c9d5e3;padding:8px;text-align:center}th{background:#eaf2fb;color:#123b63}.small{font-size:.92rem}.pill{display:inline-block;padding:2px 8px;border-radius:999px;background:#eef4fb;color:#123b63;font-weight:700}.quiz details{border:1px solid var(--line);border-radius:12px;margin:8px 0;padding:10px;background:#fff}summary{cursor:pointer;font-weight:800;color:#123b63}.svgbox{background:#fbfdff;border:1px solid var(--line);border-radius:14px;padding:10px;overflow:auto}.audio{display:flex;gap:8px;flex-wrap:wrap;align-items:center}.warn{background:#fff1f2;border-left:5px solid var(--red);padding:12px;border-radius:12px}\n@media print{nav,.audio,.buttons{display:none}.card{box-shadow:none;break-inside:avoid}body{background:white}}\n<\/style>\n<\/head><body>\n<header><h1>CH3 – Avancement d\u2019une transformation chimique<\/h1><p>Tableau d\u2019avancement, avancement maximal, r\u00e9actif limitant et composition finale. Version enrichie \u00e0 partir de la trace \u00e9crite.<\/p><\/header>\n<nav><a href=\"#cours\">Cours<\/a><a href=\"#methode\">M\u00e9thode<\/a><a href=\"#exos\">Exercices int\u00e9gr\u00e9s<\/a><a href=\"#bac\">Applications bac<\/a><a href=\"#quiz\">Quiz<\/a><a class=\"btn red\" href=\"https:\/\/pcwallis.malo.wf\/wp-content\/uploads\/2026\/06\/chapitre_avancement_premiere_cours.pdf\">PDF cours<\/a><a class=\"btn green\" href=\"https:\/\/pcwallis.malo.wf\/wp-content\/uploads\/2026\/06\/chapitre_avancement_premiere_exercices_types.pdf\">PDF exercices<\/a><\/nav>\n<main>\n<section class=\"card audio\"><strong>Lecture audio :<\/strong><button onclick=\"readPage()\">Lire le cours<\/button><button onclick=\"speechSynthesis.cancel()\">Stop<\/button><span class=\"small\">Utilise la synth\u00e8se vocale du navigateur.<\/span><\/section>\n<section id=\"cours\" class=\"card\"><h2>1. R\u00e9action chimique<\/h2><p>Une r\u00e9action chimique se produit entre des esp\u00e8ces chimiques. Les <strong>r\u00e9actifs<\/strong> sont consomm\u00e9s et les <strong>produits<\/strong> sont form\u00e9s.<\/p><div class=\"formula\">a A + b B \u2192 c C + d D<\/div><p>A et B sont les r\u00e9actifs ; C et D sont les produits ; a, b, c et d sont les coefficients st\u0153chiom\u00e9triques. Une \u00e9quation correctement ajust\u00e9e conserve les \u00e9l\u00e9ments chimiques et la charge.<\/p><div class=\"important\"><strong>Exemple de ta trace \u00e9crite :<\/strong> CH<sub>4<\/sub> + 2 O<sub>2<\/sub> \u2192 CO<sub>2<\/sub> + 2 H<sub>2<\/sub>O.<\/div><\/section>\n<section class=\"card\" id=\"methode\"><h2>2. Avancement et tableau d\u2019avancement<\/h2><p>L\u2019avancement <strong>x<\/strong> est une grandeur en mol qui permet de suivre l\u2019\u00e9volution des quantit\u00e9s de mati\u00e8re pendant une transformation.<\/p><table><tr><th>\u00c9tat<\/th><th>CH<sub>4<\/sub><\/th><th>O<sub>2<\/sub><\/th><th>CO<sub>2<\/sub><\/th><th>H<sub>2<\/sub>O<\/th><\/tr><tr><td>Initial<\/td><td>n<sub>0<\/sub>(CH<sub>4<\/sub>)<\/td><td>n<sub>0<\/sub>(O<sub>2<\/sub>)<\/td><td>0<\/td><td>0<\/td><\/tr><tr><td>En cours<\/td><td>n<sub>0<\/sub>(CH<sub>4<\/sub>) \u2212 x<\/td><td>n<sub>0<\/sub>(O<sub>2<\/sub>) \u2212 2x<\/td><td>x<\/td><td>2x<\/td><\/tr><tr><td>Final<\/td><td>n<sub>0<\/sub>(CH<sub>4<\/sub>) \u2212 x<sub>max<\/sub><\/td><td>n<sub>0<\/sub>(O<sub>2<\/sub>) \u2212 2x<sub>max<\/sub><\/td><td>x<sub>max<\/sub><\/td><td>2x<sub>max<\/sub><\/td><\/tr><\/table><div class=\"method\"><strong>Ta m\u00e9thode \u00e0 conserver :<\/strong> on calcule n<sub>0<\/sub>\/\u03bd pour chaque r\u00e9actif. Le plus petit r\u00e9sultat donne x<sub>max<\/sub> et le r\u00e9actif limitant.<\/div><div class=\"formula\">x<sub>max<\/sub> = min( n<sub>0<\/sub>(A)\/a ; n<sub>0<\/sub>(B)\/b )<\/div><\/section>\n<section class=\"card\"><h2>3. Sch\u00e9ma de raisonnement<\/h2><div class=\"svgbox\"><svg width=\"820\" height=\"230\" viewBox=\"0 0 820 230\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><defs><marker id=\"arrow\" markerWidth=\"10\" markerHeight=\"10\" refX=\"8\" refY=\"3\" orient=\"auto\"><path d=\"M0,0 L0,6 L9,3 z\" fill=\"#1d4e89\"\/><\/marker><\/defs><rect x=\"20\" y=\"55\" width=\"145\" height=\"80\" rx=\"15\" fill=\"#eaf2fb\" stroke=\"#1d4e89\"\/><text x=\"92\" y=\"88\" text-anchor=\"middle\" font-weight=\"700\">\u00c9quation<\/text><text x=\"92\" y=\"112\" text-anchor=\"middle\">ajust\u00e9e<\/text><line x1=\"165\" y1=\"95\" x2=\"235\" y2=\"95\" stroke=\"#1d4e89\" stroke-width=\"3\" marker-end=\"url(#arrow)\"\/><rect x=\"235\" y=\"55\" width=\"145\" height=\"80\" rx=\"15\" fill=\"#fff8e6\" stroke=\"#d9a441\"\/><text x=\"308\" y=\"88\" text-anchor=\"middle\" font-weight=\"700\">Quantit\u00e9s<\/text><text x=\"308\" y=\"112\" text-anchor=\"middle\">initiales<\/text><line x1=\"380\" y1=\"95\" x2=\"450\" y2=\"95\" stroke=\"#1d4e89\" stroke-width=\"3\" marker-end=\"url(#arrow)\"\/><rect x=\"450\" y=\"55\" width=\"145\" height=\"80\" rx=\"15\" fill=\"#f1f7ff\" stroke=\"#1d4e89\"\/><text x=\"523\" y=\"88\" text-anchor=\"middle\" font-weight=\"700\">Tableau<\/text><text x=\"523\" y=\"112\" text-anchor=\"middle\">d\u2019avancement<\/text><line x1=\"595\" y1=\"95\" x2=\"665\" y2=\"95\" stroke=\"#1d4e89\" stroke-width=\"3\" marker-end=\"url(#arrow)\"\/><rect x=\"665\" y=\"55\" width=\"135\" height=\"80\" rx=\"15\" fill=\"#fff1f2\" stroke=\"#9b1c31\"\/><text x=\"733\" y=\"86\" text-anchor=\"middle\" font-weight=\"700\">x<tspan baseline-shift=\"sub\">max<\/tspan><\/text><text x=\"733\" y=\"112\" text-anchor=\"middle\">\u00e9tat final<\/text><text x=\"410\" y=\"180\" text-anchor=\"middle\" fill=\"#9b1c31\" font-weight=\"700\">On ne choisit jamais le r\u00e9actif limitant au hasard : on compare n\u2080 \/ coefficient.<\/text><\/svg><\/div><\/section>\n<section id=\"exos\" class=\"card\"><h2>4. Exercice int\u00e9gr\u00e9 type cours<\/h2><p>On fait r\u00e9agir 3,0 \u00d7 10<sup>\u22122<\/sup> mol de CH<sub>4<\/sub> avec 5,0 \u00d7 10<sup>\u22122<\/sup> mol de O<sub>2<\/sub>.<\/p><details><summary>Voir la correction guid\u00e9e<\/summary><p>Pour CH<sub>4<\/sub> : x = 3,0 \u00d7 10<sup>\u22122<\/sup> \/ 1 = 3,0 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><p>Pour O<sub>2<\/sub> : x = 5,0 \u00d7 10<sup>\u22122<\/sup> \/ 2 = 2,5 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><p>Donc O<sub>2<\/sub> est le r\u00e9actif limitant et x<sub>max<\/sub> = 2,5 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><p>\u00c9tat final : CH<sub>4<\/sub> restant = 5,0 \u00d7 10<sup>\u22123<\/sup> mol ; O<sub>2<\/sub> = 0 ; CO<sub>2<\/sub> = 2,5 \u00d7 10<sup>\u22122<\/sup> mol ; H<sub>2<\/sub>O = 5,0 \u00d7 10<sup>\u22122<\/sup> mol.<\/p><\/details><\/section>\n<section id=\"bac\" class=\"card\"><h2>5. Projection bac : r\u00e9action redox et couleur finale<\/h2><p>On exploite une r\u00e9action donn\u00e9e et on applique exactement la m\u00eame m\u00e9thode d\u2019avancement.<\/p><div class=\"formula\">MnO<sub>4<\/sub><sup>\u2212<\/sup> + 5 Fe<sup>2+<\/sup> + 8 H<sup>+<\/sup> \u2192 Mn<sup>2+<\/sup> + 5 Fe<sup>3+<\/sup> + 4 H<sub>2<\/sub>O<\/div><p>Les ions MnO<sub>4<\/sub><sup>\u2212<\/sup> sont violets. Si MnO<sub>4<\/sub><sup>\u2212<\/sup> reste \u00e0 la fin, la solution reste violette. S\u2019il est totalement consomm\u00e9, la couleur violette dispara\u00eet.<\/p><div class=\"warn\"><strong>Comp\u00e9tence bac :<\/strong> d\u00e9terminer les quantit\u00e9s initiales \u00e0 partir de C \u00d7 V, construire le tableau, identifier le r\u00e9actif limitant, puis interpr\u00e9ter la couleur finale.<\/div><\/section>\n<section id=\"quiz\" class=\"card quiz\"><h2>6. Quiz de fin<\/h2><details><summary>1. Que repr\u00e9sente x ?<\/summary><p>L\u2019avancement de la r\u00e9action, exprim\u00e9 en mol.<\/p><\/details><details><summary>2. Comment trouve-t-on le r\u00e9actif limitant ?<\/summary><p>On compare n<sub>0<\/sub>\/\u03bd pour chaque r\u00e9actif. Le plus petit r\u00e9sultat correspond au r\u00e9actif limitant.<\/p><\/details><details><summary>3. Pourquoi diviser par le coefficient st\u0153chiom\u00e9trique ?<\/summary><p>Parce que la r\u00e9action consomme les r\u00e9actifs dans les proportions donn\u00e9es par l\u2019\u00e9quation ajust\u00e9e.<\/p><\/details><details><summary>4. Que permet la ligne finale du tableau ?<\/summary><p>Elle donne la composition finale du syst\u00e8me : r\u00e9actifs restants et produits form\u00e9s.<\/p><\/details><details><summary>5. Dans la r\u00e9action avec MnO<sub>4<\/sub><sup>\u2212<\/sup>, que signifie une couleur violette finale ?<\/summary><p>Il reste des ions permanganate MnO<sub>4<\/sub><sup>\u2212<\/sup> : ils n\u2019\u00e9taient pas limitants.<\/p><\/details><\/section>\n<\/main><script>function readPage(){speechSynthesis.cancel();const t=document.querySelector('main').innerText;const u=new SpeechSynthesisUtterance(t);u.lang='fr-FR';u.rate=.95;speechSynthesis.speak(u);}<\/script><\/body><\/html>\n","protected":false},"excerpt":{"rendered":"<p>CH3 – Avancement d\u2019une transformation chimique CH3 – Avancement d\u2019une transformation chimique Tableau d\u2019avancement, avancement maximal, r\u00e9actif limitant et composition finale. Version enrichie \u00e0 partir de...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-961","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/961","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=961"}],"version-history":[{"count":5,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/961\/revisions"}],"predecessor-version":[{"id":971,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/961\/revisions\/971"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=961"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}