{"id":85825,"date":"2025-06-01T03:28:53","date_gmt":"2025-06-01T03:28:53","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=85825"},"modified":"2025-06-01T03:28:53","modified_gmt":"2025-06-01T03:28:53","slug":"in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga\/","title":{"rendered":"In which one of the following reactions, the maximum quantity of H\u2082 ga"},"content":{"rendered":"<p>In which one of the following reactions, the maximum quantity of H\u2082 gas is produced by the decomposition of 1 g of compound by H\u2082O\/O\u2082?<\/p>\n<p>[amp_mcq option1=&#8221;CH\u2084 + H\u2082O \u2192 CO + 3H\u2082&#8221; option2=&#8221;CO + H\u2082O \u2192 CO\u2082 + H\u2082&#8221; option3=&#8221;CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082&#8221; option4=&#8221;C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082&#8221; correct=&#8221;option1&#8243;]<\/p>\n<div class=\"psc-box-pyq-exam-year-detail\">\n<div class=\"pyq-exam\">\n<div class=\"psc-heading\">This question was previously asked in<\/div>\n<div class=\"psc-title line-ellipsis\">UPSC CDS-2 &#8211; 2018<\/div>\n<\/div>\n<div class=\"pyq-exam-psc-buttons\"><a href=\"\/pyq\/pyq-upsc-cds-2-2018.pdf\" target=\"_blank\" class=\"psc-pdf-button\" rel=\"noopener\">Download PDF<\/a><a href=\"\/pyq-upsc-cds-2-2018\" target=\"_blank\" class=\"psc-attempt-button\" rel=\"noopener\">Attempt Online<\/a><\/div>\n<\/div>\n<section id=\"pyq-correct-answer\">\nThe reaction in option A produces the maximum quantity of H\u2082 gas per gram of the reactant compound.<br \/>\n<\/section>\n<section id=\"pyq-key-points\">\nTo find the maximum quantity of H\u2082 produced per gram of reactant, we need to calculate the moles of H\u2082 produced per gram of the principal reactant in each reaction. Let&#8217;s calculate the moles of H\u2082 per gram of the compound undergoing decomposition\/reaction:<br \/>\nA) CH\u2084 + H\u2082O \u2192 CO + 3H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 3 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (3 mol H\u2082) \/ (16 g CH\u2084) = 3\/16 mol\/g.<br \/>\nB) CO + H\u2082O \u2192 CO\u2082 + H\u2082: Molar mass of CO is 28 g\/mol. 1 mol CO produces 1 mol H\u2082. Moles of H\u2082 per gram of CO = (1 mol H\u2082) \/ (28 g CO) = 1\/28 mol\/g.<br \/>\nC) CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 2 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (2 mol H\u2082) \/ (16 g CH\u2084) = 2\/16 = 1\/8 mol\/g.<br \/>\nD) C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082: Molar mass of C\u2081\u2082H\u2082\u2084 is 168 g\/mol. 1 mol C\u2081\u2082H\u2082\u2084 produces 12 mol H\u2082. Moles of H\u2082 per gram of C\u2081\u2082H\u2082\u2084 = (12 mol H\u2082) \/ (168 g C\u2081\u2082H\u2082\u2084) = 12\/168 = 1\/14 mol\/g.<br \/>\nComparing the values: 3\/16 (0.1875), 1\/28 (~0.0357), 1\/8 (0.125), 1\/14 (~0.0714). The largest value is 3\/16 mol\/g, corresponding to reaction A.<br \/>\n<\/section>\n<section id=\"pyq-additional-information\">\nThese reactions represent different methods for producing hydrogen gas, often from hydrocarbon fuels or syngas (CO + H\u2082). Reaction A is a simplified steam reforming of methane. Reaction B is the water-gas shift reaction. Reaction C is partial oxidation of methane. Reaction D involves a larger hydrocarbon. The amount of hydrogen produced depends on the stoichiometry of the reaction and the molar mass of the starting material.<br \/>\n<\/section>\n","protected":false},"excerpt":{"rendered":"<p>In which one of the following reactions, the maximum quantity of H\u2082 gas is produced by the decomposition of 1 g of compound by H\u2082O\/O\u2082? [amp_mcq option1=&#8221;CH\u2084 + H\u2082O \u2192 CO + 3H\u2082&#8221; option2=&#8221;CO + H\u2082O \u2192 CO\u2082 + H\u2082&#8221; option3=&#8221;CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082&#8221; option4=&#8221;C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082&#8221; correct=&#8221;option1&#8243;] &#8230; <\/p>\n<p class=\"read-more-container\"><a title=\"In which one of the following reactions, the maximum quantity of H\u2082 ga\" class=\"read-more button\" href=\"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga\/#more-85825\">Detailed Solution<span class=\"screen-reader-text\">In which one of the following reactions, the maximum quantity of H\u2082 ga<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1088],"tags":[1114,1096,1330],"class_list":["post-85825","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-2","tag-1114","tag-chemistry","tag-hydrocarbons","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>In which one of the following reactions, the maximum quantity of H\u2082 ga<\/title>\n<meta name=\"description\" content=\"The reaction in option A produces the maximum quantity of H\u2082 gas per gram of the reactant compound. To find the maximum quantity of H\u2082 produced per gram of reactant, we need to calculate the moles of H\u2082 produced per gram of the principal reactant in each reaction. Let&#039;s calculate the moles of H\u2082 per gram of the compound undergoing decomposition\/reaction: A) CH\u2084 + H\u2082O \u2192 CO + 3H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 3 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (3 mol H\u2082) \/ (16 g CH\u2084) = 3\/16 mol\/g. B) CO + H\u2082O \u2192 CO\u2082 + H\u2082: Molar mass of CO is 28 g\/mol. 1 mol CO produces 1 mol H\u2082. Moles of H\u2082 per gram of CO = (1 mol H\u2082) \/ (28 g CO) = 1\/28 mol\/g. C) CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 2 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (2 mol H\u2082) \/ (16 g CH\u2084) = 2\/16 = 1\/8 mol\/g. D) C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082: Molar mass of C\u2081\u2082H\u2082\u2084 is 168 g\/mol. 1 mol C\u2081\u2082H\u2082\u2084 produces 12 mol H\u2082. Moles of H\u2082 per gram of C\u2081\u2082H\u2082\u2084 = (12 mol H\u2082) \/ (168 g C\u2081\u2082H\u2082\u2084) = 12\/168 = 1\/14 mol\/g. Comparing the values: 3\/16 (0.1875), 1\/28 (~0.0357), 1\/8 (0.125), 1\/14 (~0.0714). The largest value is 3\/16 mol\/g, corresponding to reaction A.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h\u2082-ga\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In which one of the following reactions, the maximum quantity of H\u2082 ga\" \/>\n<meta property=\"og:description\" content=\"The reaction in option A produces the maximum quantity of H\u2082 gas per gram of the reactant compound. To find the maximum quantity of H\u2082 produced per gram of reactant, we need to calculate the moles of H\u2082 produced per gram of the principal reactant in each reaction. Let&#039;s calculate the moles of H\u2082 per gram of the compound undergoing decomposition\/reaction: A) CH\u2084 + H\u2082O \u2192 CO + 3H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 3 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (3 mol H\u2082) \/ (16 g CH\u2084) = 3\/16 mol\/g. B) CO + H\u2082O \u2192 CO\u2082 + H\u2082: Molar mass of CO is 28 g\/mol. 1 mol CO produces 1 mol H\u2082. Moles of H\u2082 per gram of CO = (1 mol H\u2082) \/ (28 g CO) = 1\/28 mol\/g. C) CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 2 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (2 mol H\u2082) \/ (16 g CH\u2084) = 2\/16 = 1\/8 mol\/g. D) C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082: Molar mass of C\u2081\u2082H\u2082\u2084 is 168 g\/mol. 1 mol C\u2081\u2082H\u2082\u2084 produces 12 mol H\u2082. Moles of H\u2082 per gram of C\u2081\u2082H\u2082\u2084 = (12 mol H\u2082) \/ (168 g C\u2081\u2082H\u2082\u2084) = 12\/168 = 1\/14 mol\/g. Comparing the values: 3\/16 (0.1875), 1\/28 (~0.0357), 1\/8 (0.125), 1\/14 (~0.0714). The largest value is 3\/16 mol\/g, corresponding to reaction A.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h\u2082-ga\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T03:28:53+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"In which one of the following reactions, the maximum quantity of H\u2082 ga","description":"The reaction in option A produces the maximum quantity of H\u2082 gas per gram of the reactant compound. To find the maximum quantity of H\u2082 produced per gram of reactant, we need to calculate the moles of H\u2082 produced per gram of the principal reactant in each reaction. Let's calculate the moles of H\u2082 per gram of the compound undergoing decomposition\/reaction: A) CH\u2084 + H\u2082O \u2192 CO + 3H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 3 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (3 mol H\u2082) \/ (16 g CH\u2084) = 3\/16 mol\/g. B) CO + H\u2082O \u2192 CO\u2082 + H\u2082: Molar mass of CO is 28 g\/mol. 1 mol CO produces 1 mol H\u2082. Moles of H\u2082 per gram of CO = (1 mol H\u2082) \/ (28 g CO) = 1\/28 mol\/g. C) CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 2 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (2 mol H\u2082) \/ (16 g CH\u2084) = 2\/16 = 1\/8 mol\/g. D) C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082: Molar mass of C\u2081\u2082H\u2082\u2084 is 168 g\/mol. 1 mol C\u2081\u2082H\u2082\u2084 produces 12 mol H\u2082. Moles of H\u2082 per gram of C\u2081\u2082H\u2082\u2084 = (12 mol H\u2082) \/ (168 g C\u2081\u2082H\u2082\u2084) = 12\/168 = 1\/14 mol\/g. Comparing the values: 3\/16 (0.1875), 1\/28 (~0.0357), 1\/8 (0.125), 1\/14 (~0.0714). The largest value is 3\/16 mol\/g, corresponding to reaction A.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h\u2082-ga\/","og_locale":"en_US","og_type":"article","og_title":"In which one of the following reactions, the maximum quantity of H\u2082 ga","og_description":"The reaction in option A produces the maximum quantity of H\u2082 gas per gram of the reactant compound. To find the maximum quantity of H\u2082 produced per gram of reactant, we need to calculate the moles of H\u2082 produced per gram of the principal reactant in each reaction. Let's calculate the moles of H\u2082 per gram of the compound undergoing decomposition\/reaction: A) CH\u2084 + H\u2082O \u2192 CO + 3H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 3 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (3 mol H\u2082) \/ (16 g CH\u2084) = 3\/16 mol\/g. B) CO + H\u2082O \u2192 CO\u2082 + H\u2082: Molar mass of CO is 28 g\/mol. 1 mol CO produces 1 mol H\u2082. Moles of H\u2082 per gram of CO = (1 mol H\u2082) \/ (28 g CO) = 1\/28 mol\/g. C) CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 2 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (2 mol H\u2082) \/ (16 g CH\u2084) = 2\/16 = 1\/8 mol\/g. D) C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082: Molar mass of C\u2081\u2082H\u2082\u2084 is 168 g\/mol. 1 mol C\u2081\u2082H\u2082\u2084 produces 12 mol H\u2082. Moles of H\u2082 per gram of C\u2081\u2082H\u2082\u2084 = (12 mol H\u2082) \/ (168 g C\u2081\u2082H\u2082\u2084) = 12\/168 = 1\/14 mol\/g. Comparing the values: 3\/16 (0.1875), 1\/28 (~0.0357), 1\/8 (0.125), 1\/14 (~0.0714). The largest value is 3\/16 mol\/g, corresponding to reaction A.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h\u2082-ga\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T03:28:53+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga\/","url":"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga\/","name":"In which one of the following reactions, the maximum quantity of H\u2082 ga","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T03:28:53+00:00","dateModified":"2025-06-01T03:28:53+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The reaction in option A produces the maximum quantity of H\u2082 gas per gram of the reactant compound. To find the maximum quantity of H\u2082 produced per gram of reactant, we need to calculate the moles of H\u2082 produced per gram of the principal reactant in each reaction. Let's calculate the moles of H\u2082 per gram of the compound undergoing decomposition\/reaction: A) CH\u2084 + H\u2082O \u2192 CO + 3H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 3 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (3 mol H\u2082) \/ (16 g CH\u2084) = 3\/16 mol\/g. B) CO + H\u2082O \u2192 CO\u2082 + H\u2082: Molar mass of CO is 28 g\/mol. 1 mol CO produces 1 mol H\u2082. Moles of H\u2082 per gram of CO = (1 mol H\u2082) \/ (28 g CO) = 1\/28 mol\/g. C) CH\u2084 + \u00bdO\u2082 \u2192 CO + 2H\u2082: Molar mass of CH\u2084 is 16 g\/mol. 1 mol CH\u2084 produces 2 mol H\u2082. Moles of H\u2082 per gram of CH\u2084 = (2 mol H\u2082) \/ (16 g CH\u2084) = 2\/16 = 1\/8 mol\/g. D) C\u2081\u2082H\u2082\u2084 + 6O\u2082 \u2192 12CO + 12H\u2082: Molar mass of C\u2081\u2082H\u2082\u2084 is 168 g\/mol. 1 mol C\u2081\u2082H\u2082\u2084 produces 12 mol H\u2082. Moles of H\u2082 per gram of C\u2081\u2082H\u2082\u2084 = (12 mol H\u2082) \/ (168 g C\u2081\u2082H\u2082\u2084) = 12\/168 = 1\/14 mol\/g. Comparing the values: 3\/16 (0.1875), 1\/28 (~0.0357), 1\/8 (0.125), 1\/14 (~0.0714). The largest value is 3\/16 mol\/g, corresponding to reaction A.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/in-which-one-of-the-following-reactions-the-maximum-quantity-of-h%e2%82%82-ga\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-2\/"},{"@type":"ListItem","position":3,"name":"In which one of the following reactions, the maximum quantity of H\u2082 ga"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/85825","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=85825"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/85825\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=85825"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=85825"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=85825"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}