{"id":92801,"date":"2025-06-01T11:32:52","date_gmt":"2025-06-01T11:32:52","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=92801"},"modified":"2025-06-01T11:32:52","modified_gmt":"2025-06-01T11:32:52","slug":"manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/","title":{"rendered":"Manganese is extracted from Manganese dioxide by reaction with Alumini"},"content":{"rendered":"

Manganese is extracted from Manganese dioxide by reaction with Aluminium as described by the following unbalanced chemical equation :
\nMnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s)
\nThe number of moles of Al (s) required to form one mole of Mn from its oxide is<\/p>\n

[amp_mcq option1=”1″ option2=”0.75″ option3=”1.33″ option4=”2″ correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CISF-AC-EXE – 2020<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe unbalanced chemical equation is MnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s). To determine the moles of Al required to form one mole of Mn, we first need to balance the equation.
\nBalancing the oxygen atoms (2 on the left, 3 on the right), we find the least common multiple is 6. Multiply MnO\u2082 by 3 and Al\u2082O\u2083 by 2:
\n3 MnO\u2082(s) + Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s)
\nNow, balance the aluminium atoms (1 on the left, 4 on the right). Multiply Al by 4:
\n3 MnO\u2082(s) + 4 Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s)
\nFinally, balance the manganese atoms (3 on the left, 1 on the right). Multiply Mn by 3:
\n3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s)
\nThe balanced equation is 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s).
\nAccording to the balanced equation, 4 moles of Al react to produce 3 moles of Mn. To produce 1 mole of Mn, the number of moles of Al required is (4 moles Al \/ 3 moles Mn) * 1 mole Mn = 4\/3 moles Al.
\n<\/section>\n
\nThe question requires balancing the given chemical equation and then using stoichiometry to find the mole ratio between reactants and products. The ratio of Al to Mn in the balanced equation is 4:3.
\n<\/section>\n
\nThis reaction is a type of redox reaction, specifically a thermite reaction where a metal oxide is reduced by a more reactive metal (Aluminium). Aluminium is a strong reducing agent. The value 4\/3 is approximately 1.333…
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Manganese is extracted from Manganese dioxide by reaction with Aluminium as described by the following unbalanced chemical equation : MnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s) The number of moles of Al (s) required to form one mole of Mn from its oxide is [amp_mcq option1=”1″ option2=”0.75″ option3=”1.33″ option4=”2″ correct=”option3″] This question … <\/p>\n

Detailed SolutionManganese is extracted from Manganese dioxide by reaction with Alumini<\/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":[1089],"tags":[1288,1096,1109],"class_list":["post-92801","post","type-post","status-publish","format-standard","hentry","category-upsc-cisf-ac-exe","tag-1288","tag-chemistry","tag-metals-minerals-ores-properties-uses","no-featured-image-padding"],"yoast_head":"\nManganese is extracted from Manganese dioxide by reaction with Alumini<\/title>\n<meta name=\"description\" content=\"The unbalanced chemical equation is MnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s). To determine the moles of Al required to form one mole of Mn, we first need to balance the equation. Balancing the oxygen atoms (2 on the left, 3 on the right), we find the least common multiple is 6. Multiply MnO\u2082 by 3 and Al\u2082O\u2083 by 2: 3 MnO\u2082(s) + Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Now, balance the aluminium atoms (1 on the left, 4 on the right). Multiply Al by 4: 3 MnO\u2082(s) + 4 Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Finally, balance the manganese atoms (3 on the left, 1 on the right). Multiply Mn by 3: 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s) The balanced equation is 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s). According to the balanced equation, 4 moles of Al react to produce 3 moles of Mn. To produce 1 mole of Mn, the number of moles of Al required is (4 moles Al \/ 3 moles Mn) * 1 mole Mn = 4\/3 moles Al. The question requires balancing the given chemical equation and then using stoichiometry to find the mole ratio between reactants and products. The ratio of Al to Mn in the balanced equation is 4:3.\" \/>\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\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Manganese is extracted from Manganese dioxide by reaction with Alumini\" \/>\n<meta property=\"og:description\" content=\"The unbalanced chemical equation is MnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s). To determine the moles of Al required to form one mole of Mn, we first need to balance the equation. Balancing the oxygen atoms (2 on the left, 3 on the right), we find the least common multiple is 6. Multiply MnO\u2082 by 3 and Al\u2082O\u2083 by 2: 3 MnO\u2082(s) + Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Now, balance the aluminium atoms (1 on the left, 4 on the right). Multiply Al by 4: 3 MnO\u2082(s) + 4 Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Finally, balance the manganese atoms (3 on the left, 1 on the right). Multiply Mn by 3: 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s) The balanced equation is 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s). According to the balanced equation, 4 moles of Al react to produce 3 moles of Mn. To produce 1 mole of Mn, the number of moles of Al required is (4 moles Al \/ 3 moles Mn) * 1 mole Mn = 4\/3 moles Al. The question requires balancing the given chemical equation and then using stoichiometry to find the mole ratio between reactants and products. The ratio of Al to Mn in the balanced equation is 4:3.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:32:52+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=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Manganese is extracted from Manganese dioxide by reaction with Alumini","description":"The unbalanced chemical equation is MnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s). To determine the moles of Al required to form one mole of Mn, we first need to balance the equation. Balancing the oxygen atoms (2 on the left, 3 on the right), we find the least common multiple is 6. Multiply MnO\u2082 by 3 and Al\u2082O\u2083 by 2: 3 MnO\u2082(s) + Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Now, balance the aluminium atoms (1 on the left, 4 on the right). Multiply Al by 4: 3 MnO\u2082(s) + 4 Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Finally, balance the manganese atoms (3 on the left, 1 on the right). Multiply Mn by 3: 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s) The balanced equation is 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s). According to the balanced equation, 4 moles of Al react to produce 3 moles of Mn. To produce 1 mole of Mn, the number of moles of Al required is (4 moles Al \/ 3 moles Mn) * 1 mole Mn = 4\/3 moles Al. The question requires balancing the given chemical equation and then using stoichiometry to find the mole ratio between reactants and products. The ratio of Al to Mn in the balanced equation is 4:3.","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\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/","og_locale":"en_US","og_type":"article","og_title":"Manganese is extracted from Manganese dioxide by reaction with Alumini","og_description":"The unbalanced chemical equation is MnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s). To determine the moles of Al required to form one mole of Mn, we first need to balance the equation. Balancing the oxygen atoms (2 on the left, 3 on the right), we find the least common multiple is 6. Multiply MnO\u2082 by 3 and Al\u2082O\u2083 by 2: 3 MnO\u2082(s) + Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Now, balance the aluminium atoms (1 on the left, 4 on the right). Multiply Al by 4: 3 MnO\u2082(s) + 4 Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Finally, balance the manganese atoms (3 on the left, 1 on the right). Multiply Mn by 3: 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s) The balanced equation is 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s). According to the balanced equation, 4 moles of Al react to produce 3 moles of Mn. To produce 1 mole of Mn, the number of moles of Al required is (4 moles Al \/ 3 moles Mn) * 1 mole Mn = 4\/3 moles Al. The question requires balancing the given chemical equation and then using stoichiometry to find the mole ratio between reactants and products. The ratio of Al to Mn in the balanced equation is 4:3.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:32:52+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/","url":"https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/","name":"Manganese is extracted from Manganese dioxide by reaction with Alumini","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:32:52+00:00","dateModified":"2025-06-01T11:32:52+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The unbalanced chemical equation is MnO\u2082(s) + Al (s) \u2192 Mn (l) + Al\u2082O\u2083 (s). To determine the moles of Al required to form one mole of Mn, we first need to balance the equation. Balancing the oxygen atoms (2 on the left, 3 on the right), we find the least common multiple is 6. Multiply MnO\u2082 by 3 and Al\u2082O\u2083 by 2: 3 MnO\u2082(s) + Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Now, balance the aluminium atoms (1 on the left, 4 on the right). Multiply Al by 4: 3 MnO\u2082(s) + 4 Al (s) \u2192 Mn (l) + 2 Al\u2082O\u2083 (s) Finally, balance the manganese atoms (3 on the left, 1 on the right). Multiply Mn by 3: 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s) The balanced equation is 3 MnO\u2082(s) + 4 Al (s) \u2192 3 Mn (l) + 2 Al\u2082O\u2083 (s). According to the balanced equation, 4 moles of Al react to produce 3 moles of Mn. To produce 1 mole of Mn, the number of moles of Al required is (4 moles Al \/ 3 moles Mn) * 1 mole Mn = 4\/3 moles Al. The question requires balancing the given chemical equation and then using stoichiometry to find the mole ratio between reactants and products. The ratio of Al to Mn in the balanced equation is 4:3.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/manganese-is-extracted-from-manganese-dioxide-by-reaction-with-alumini\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CISF-AC-EXE","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cisf-ac-exe\/"},{"@type":"ListItem","position":3,"name":"Manganese is extracted from Manganese dioxide by reaction with Alumini"}]},{"@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\/92801","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=92801"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/92801\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=92801"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=92801"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=92801"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}