To solve this problem, we use stoichiometry and the concept of percent yield<\/strong>.<\/p>\n\n\n\n $$ This equation tells us that 2 moles of lithium (Li)<\/strong> react to form 1 mole of hydrogen gas (H\u2082)<\/strong>.<\/p>\n\n\n\n You are told that 0.30 mol<\/strong> of H\u2082 was isolated with an 85% yield<\/strong>. This means that 0.30 mol is only 85%<\/strong> of the amount of H\u2082 that would have been produced if the reaction had gone to 100% completion (theoretical yield)<\/strong>.<\/p>\n\n\n\n Use the percent yield formula:<\/p>\n\n\n\n $$ Rearrange to solve for theoretical yield:<\/p>\n\n\n\n $$ From the balanced equation:<\/p>\n\n\n\n $$ So to make 0.3529 mol of H\u2082<\/strong>, we need:<\/p>\n\n\n\n $$ $$ This problem involves applying basic stoichiometry and yield concepts to a chemical reaction. Lithium reacts with water to produce lithium hydroxide and hydrogen gas. The balanced equation, $2Li + 2H_2O \\rightarrow 2LiOH + H_2$, shows the stoichiometric relationship between reactants and products. For every 2 moles of lithium<\/strong>, 1 mole of hydrogen gas<\/strong> is formed. This 2:1 molar ratio is key for the conversion.<\/p>\n\n\n\n However, the reaction did not proceed with 100% efficiency. Instead, the reaction yield was 85%, meaning the amount of hydrogen gas actually collected (0.30 mol) is less than what would be predicted if all lithium reacted perfectly. This necessitates calculating the theoretical yield<\/strong>, the amount of hydrogen that would be produced with 100% efficiency, using the formula:<\/p>\n\n\n\n $$ Plugging in the numbers: $0.30 \\div 0.85 \\approx 0.3529$ mol H\u2082. This is the amount of hydrogen that would be formed if the reaction had gone to completion with no loss.<\/p>\n\n\n\n Next, using the stoichiometric ratio, we find the amount of lithium required to produce that theoretical amount of H\u2082. Since 2 mol of Li make 1 mol of H\u2082, for 0.3529 mol of H\u2082 we need $2 \\times 0.3529 = 0.7058$ mol of Li.<\/p>\n\n\n\n After rounding appropriately, the final answer is 0.71 mol of lithium<\/strong> reacted. This approach ensures accurate, chemistry-based reasoning to solve real-world laboratory problems<\/p>\n\n\n\n .<\/p>\n\n\n\n The reaction of lithium metal and water to form lithium hydroxide and hydrogen gas is represented by the following balanced equation: 2Li(s) + 2H2O(l) \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 2LiOH(aq) + H2(g) When Li is mixed with excess water, 0.30 mol H2 gas is isolated in the laboratory. If this reaction occurs to give an 85% yield of […]<\/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-217693","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217693","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=217693"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217693\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217693"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217693"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217693"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Write the balanced chemical equation<\/strong><\/h3>\n\n\n\n
2Li(s) + 2H_2O(l) \\rightarrow 2LiOH(aq) + H_2(g)
$$<\/p>\n\n\n\n
\n\n\n\nStep 2: Understand percent yield<\/strong><\/h3>\n\n\n\n
\\text{Percent yield} = \\left( \\frac{\\text{Actual yield}}{\\text{Theoretical yield}} \\right) \\times 100
$$<\/p>\n\n\n\n
\\text{Theoretical yield} = \\frac{\\text{Actual yield}}{\\text{Percent yield}} = \\frac{0.30 \\, \\text{mol}}{0.85} \\approx 0.3529 \\, \\text{mol of } H_2
$$<\/p>\n\n\n\n
\n\n\n\nStep 3: Use stoichiometry to find moles of Li<\/strong><\/h3>\n\n\n\n
2 \\, \\text{mol Li} : 1 \\, \\text{mol } H_2
$$<\/p>\n\n\n\n
2 \\times 0.3529 = 0.7058 \\, \\text{mol of Li}
$$<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\\boxed{0.71 \\, \\text{mol of Li (rounded to two decimal places)}}
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (300+ words):<\/strong><\/h3>\n\n\n\n
\\text{Theoretical yield} = \\frac{\\text{Actual yield}}{\\text{Percent yield}}
$$<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-43.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"