Determine the theoretical yield in H2S (in moles) of 16 mol Al2S3 and 16 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17g\/mol.<\/p>\n\n\n\n
Al2S3(s) + 6H2O(l) –> 2Al(OH)3(s) + 3H2S(g)<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the theoretical yield of hydrogen sulfide (H\u2082S) in moles when 16 moles of aluminum sulfide (Al\u2082S\u2083) and 16 moles of water (H\u2082O) are reacted, we first need to focus on the stoichiometric relationships given in the balanced chemical equation:<\/p>\n\n\n\n $$ From the balanced equation, the mole ratio between Al\u2082S\u2083 and H\u2082S is as follows:<\/p>\n\n\n\n Similarly, the mole ratio between H\u2082O and H\u2082S is:<\/p>\n\n\n\n We are given 16 moles of Al\u2082S\u2083 and 16 moles of H\u2082O. To identify the limiting reactant, we need to compare the amount of H\u2082S that can be produced by each reactant.<\/p>\n\n\n\n The mole ratio of Al\u2082S\u2083 to H\u2082S is 1:3. Therefore, 16 moles of Al\u2082S\u2083 will produce:<\/p>\n\n\n\n $$ The mole ratio of H\u2082O to H\u2082S is 6:3 (or 2:1). Therefore, 16 moles of H\u2082O will produce:<\/p>\n\n\n\n $$ The amount of H\u2082S produced by Al\u2082S\u2083 (48 moles) is greater than the amount produced by H\u2082O (8 moles). Therefore, H\u2082O is the limiting reactant.<\/p>\n\n\n\n Since H\u2082O is the limiting reactant, the theoretical yield of H\u2082S is determined by the amount of H\u2082O available. From the calculations above, 16 moles of H\u2082O will produce 8 moles of H\u2082S.<\/p>\n\n\n\n Thus, the theoretical yield of H\u2082S is 8 moles<\/strong>.<\/p>\n\n\n\n The theoretical yield of H\u2082S when 16 moles of Al\u2082S\u2083 and 16 moles of H\u2082O react is 8 moles<\/strong>. This is because H\u2082O is the limiting reactant, and its availability dictates the maximum amount of H\u2082S that can be produced.<\/p>\n","protected":false},"excerpt":{"rendered":" Determine the theoretical yield in H2S (in moles) of 16 mol Al2S3 and 16 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17g\/mol. Al2S3(s) + 6H2O(l) –> 2Al(OH)3(s) + 3H2S(g) The correct answer and explanation is : To determine the theoretical yield of hydrogen sulfide […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-211855","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211855","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=211855"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211855\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211855"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211855"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211855"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{Al}_2\\text{S}_3(s) + 6\\text{H}_2\\text{O}(l) \\rightarrow 2\\text{Al(OH)}_3(s) + 3\\text{H}_2\\text{S}(g)
$$<\/p>\n\n\n\nStep 1: Analyze the Stoichiometric Ratios<\/h3>\n\n\n\n
\n
\n
Step 2: Determine the Limiting Reactant<\/h3>\n\n\n\n
For Al\u2082S\u2083:<\/h4>\n\n\n\n
16 \\, \\text{mol Al}_2\\text{S}_3 \\times \\frac{3 \\, \\text{mol H}_2\\text{S}}{1 \\, \\text{mol Al}_2\\text{S}_3} = 48 \\, \\text{mol H}_2\\text{S}
$$<\/p>\n\n\n\nFor H\u2082O:<\/h4>\n\n\n\n
16 \\, \\text{mol H}_2\\text{O} \\times \\frac{3 \\, \\text{mol H}_2\\text{S}}{6 \\, \\text{mol H}_2\\text{O}} = 8 \\, \\text{mol H}_2\\text{S}
$$<\/p>\n\n\n\nStep 3: Identify the Limiting Reactant<\/h3>\n\n\n\n
Step 4: Calculate the Theoretical Yield of H\u2082S<\/h3>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n