Stoichiometry: Mole To Mass Problems 1. Mg + 2HCl \u2192 MgCl2 + H2 A. How Many Grams Of HCl Are Consumed By The Reaction Of 3 Moles Of Magnesium? B. What Is The Mass In Grams Of H, Gas When 5.0 Moles Of HCl Is Added To The Reaction?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n [ From the balanced chemical equation, 1 mole of Mg reacts with 2 moles of HCl<\/strong>.<\/p>\n\n\n\n [ The molar mass of HCl is: [ Answer: 218.76 g of HCl<\/strong><\/p>\n\n\n\n From the balanced reaction: [ Molar mass of H\u2082: [ Answer: 5.05 g of H\u2082<\/strong><\/p>\n\n\n\n Stoichiometry is the study of quantitative relationships in chemical reactions. It helps us determine how much reactant is needed or how much product is formed. In this case, we use stoichiometry to convert between moles and mass.<\/p>\n\n\n\n For the first problem, we start with 3 moles of Mg and use the balanced equation to find the amount of HCl required. The reaction shows that 1 mole of Mg reacts with 2 moles of HCl, so 3 moles of Mg will need 6 moles of HCl. Then, using the molar mass of HCl (36.46 g\/mol), we convert the moles into grams, obtaining 218.76 g<\/strong>.<\/p>\n\n\n\n For the second problem, we determine how much hydrogen gas is produced when 5.0 moles of HCl reacts. According to the reaction, 2 moles of HCl produce 1 mole of H\u2082 gas. Using this mole ratio, 5.0 moles of HCl will produce 2.5 moles<\/strong> of H\u2082. Converting moles of H\u2082 to grams using its molar mass (2.02 g\/mol), we get 5.05 g<\/strong>.<\/p>\n\n\n\n This method works for any stoichiometry problem by following these steps:<\/p>\n\n\n\n Stoichiometry is essential in chemistry as it helps predict reaction outcomes, ensuring the correct amount of reactants is used to avoid waste and optimize reactions.<\/p>\n","protected":false},"excerpt":{"rendered":" Stoichiometry: Mole To Mass Problems 1. Mg + 2HCl \u2192 MgCl2 + H2 A. How Many Grams Of HCl Are Consumed By The Reaction Of 3 Moles Of Magnesium? B. What Is The Mass In Grams Of H, Gas When 5.0 Moles Of HCl Is Added To The Reaction? The Correct Answer and Explanation is […]<\/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-187474","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187474","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=187474"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187474\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187474"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187474"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187474"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Stoichiometry: Mole to Mass Problems<\/strong><\/h3>\n\n\n\n
Given Reaction:<\/strong><\/h4>\n\n\n\n
\\text{Mg} + 2\\text{HCl} \\rightarrow \\text{MgCl}_2 + \\text{H}_2
]<\/p>\n\n\n\nA. How Many Grams of HCl Are Consumed by the Reaction of 3 Moles of Magnesium?<\/strong><\/h3>\n\n\n\n
Step 1: Identify the Mole Ratio<\/strong><\/h4>\n\n\n\n
Step 2: Calculate Moles of HCl Needed<\/strong><\/h4>\n\n\n\n
3 \\text{ moles Mg} \\times \\frac{2 \\text{ moles HCl}}{1 \\text{ mole Mg}} = 6 \\text{ moles HCl}
]<\/p>\n\n\n\nStep 3: Convert Moles of HCl to Mass<\/strong><\/h4>\n\n\n\n
[
\\text{HCl} = 1.01 + 35.45 = 36.46 \\text{ g\/mol}
]<\/p>\n\n\n\n
6 \\text{ moles HCl} \\times 36.46 \\text{ g\/mol} = 218.76 \\text{ g HCl}
]<\/p>\n\n\n\n
\n\n\n\nB. What Is the Mass in Grams of H\u2082 Gas When 5.0 Moles of HCl Is Added?<\/strong><\/h3>\n\n\n\n
Step 1: Identify the Mole Ratio<\/strong><\/h4>\n\n\n\n
[
2 \\text{ moles HCl} \\rightarrow 1 \\text{ mole H}_2
]<\/p>\n\n\n\nStep 2: Calculate Moles of H\u2082 Formed<\/strong><\/h4>\n\n\n\n
5.0 \\text{ moles HCl} \\times \\frac{1 \\text{ mole H}_2}{2 \\text{ moles HCl}} = 2.5 \\text{ moles H}_2
]<\/p>\n\n\n\nStep 3: Convert Moles of H\u2082 to Mass<\/strong><\/h4>\n\n\n\n
[
2.02 \\text{ g\/mol}
]<\/p>\n\n\n\n
2.5 \\text{ moles H}_2 \\times 2.02 \\text{ g\/mol} = 5.05 \\text{ g H}_2
]<\/p>\n\n\n\n
\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n
\n