Corrected Answer<\/h3>\n\n\n\n
a. Balanced Reaction<\/strong> b. Final State of the Reaction<\/strong> This problem is an example of a limiting reactant calculation in stoichiometry. The amounts of all substances after the reaction depend on which reactant is completely consumed first. The process involves converting the initial masses of reactants into moles, using the balanced chemical equation to determine the limiting reactant, and then calculating the quantities of products formed and excess reactant remaining.<\/p>\n\n\n\n Step 1: Balancing the Chemical Equation<\/strong> Step 2: Calculating Initial Moles of Reactants<\/strong> Moles of Zn = 3.50 g Zn \/ 65.38 g\/mol = 0.0535 mol Zn Step 3: Identifying the Limiting Reactant<\/strong> Since hydrochloric acid produces a smaller amount of hydrogen gas (0.0343 mol), HCl is the limiting reactant<\/strong>. It will be completely consumed, and the reaction will stop once all the HCl has reacted.<\/p>\n\n\n\n Step 4: Calculating Final Quantities<\/strong> In this problem, zinc reacts with hydrochloric acid to make zinc chloride and hydrogen gas according to the following reaction: Zn(s) + HCl(aq) \u00e2\u2020\u2019 ZnCl2(aq) + H2(g) a. Balance the reaction b. A 3.50 g sample of zinc metal is allowed to react with 2.50 g of hydrochloric acid. Show your work below and complete […]<\/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-231187","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231187","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=231187"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231187\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231187"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231187"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231187"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Zn(s) + 2 HCl(aq) \u2192 ZnCl\u2082(aq) + H\u2082(g)<\/p>\n\n\n\n
The following table shows the correct quantities of reactants and products before and after the reaction is complete.<\/p>\n\n\n\nReactants and Products<\/td> Before Reaction<\/td> After Reaction<\/td><\/tr> Zn (grams)<\/strong><\/td> 3.50 g<\/td> 1.26 g<\/td><\/tr> HCl (grams)<\/strong><\/td> 2.50 g<\/td> 0 g<\/td><\/tr> ZnCl\u2082 (grams)<\/strong><\/td> 0 g<\/td> 4.67 g<\/td><\/tr> H\u2082 (L)<\/strong><\/td> 0 L<\/td> 0.768 L<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nExplanation<\/h3>\n\n\n\n
The initial unbalanced equation is Zn + HCl \u2192 ZnCl\u2082 + H\u2082. To balance the equation, the number of atoms of each element must be the same on both the reactant and product sides. There is one zinc atom on both sides. However, there is one hydrogen and one chlorine atom on the reactant side, but two of each on the product side. Placing a coefficient of 2 in front of HCl balances the hydrogen and chlorine atoms.
The balanced equation is: Zn(s) + 2 HCl(aq) \u2192 ZnCl\u2082(aq) + H\u2082(g)<\/strong>
This equation shows that one mole of solid zinc reacts with two moles of hydrochloric acid to produce one mole of zinc chloride and one mole of hydrogen gas.<\/p>\n\n\n\n
To determine the limiting reactant, the initial masses must be converted to moles using their respective molar masses.<\/p>\n\n\n\n\n
Moles of HCl = 2.50 g HCl \/ 36.46 g\/mol = 0.0686 mol HCl<\/p>\n\n\n\n
The limiting reactant is the one that produces the least amount of product. The amount of product formed from each reactant can be calculated using the mole ratios from the balanced equation. Let’s calculate the moles of H\u2082 that could be produced from each reactant.<\/p>\n\n\n\n\n
All subsequent calculations are based on the amount of the limiting reactant (0.0686 mol HCl).<\/p>\n\n\n\n\n
First, find the moles of ZnCl\u2082 produced.
Moles of ZnCl\u2082 = 0.0686 mol HCl \u00d7 (1 mol ZnCl\u2082 \/ 2 mol HCl) = 0.0343 mol ZnCl\u2082
Next, convert moles to grams using the molar mass of ZnCl\u2082 (136.28\u00a0g\/mol\u00a0).
Mass of ZnCl\u2082 = 0.0343 mol \u00d7 136.28\u00a0g\/mol\u00a0=\u00a04.67 g ZnCl\u2082<\/strong><\/li>\n\n\n\n
The moles of H\u2082 produced are 0.0343 mol. Assuming Standard Temperature and Pressure (STP), one mole of any ideal gas occupies 22.4 liters.
Volume of H\u2082 = 0.0343 mol \u00d7 22.4 L\/mol =\u00a00.768 L H\u2082<\/strong><\/li>\n\n\n\n
First, calculate the mass of Zn that reacted with the limiting HCl.
Moles of Zn reacted = 0.0686 mol HCl \u00d7 (1 mol Zn \/ 2 mol HCl) = 0.0343 mol Zn
Mass of Zn reacted = 0.0343 mol \u00d7 65.38\u00a0g\/mol\u00a0= 2.24 g Zn
Finally, subtract the reacted mass from the initial mass.
Mass of Zn remaining = 3.50 g (initial) – 2.24 g (reacted) =\u00a01.26 g Zn<\/strong><\/li>\n\n\n\n
Since HCl is the limiting reactant, it is completely consumed.
Mass of HCl remaining =\u00a00 g<\/strong><\/li>\n<\/ul>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-850.jpeg\")
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