{"id":239490,"date":"2025-07-02T21:15:26","date_gmt":"2025-07-02T21:15:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239490"},"modified":"2025-07-02T21:15:29","modified_gmt":"2025-07-02T21:15:29","slug":"zinc-metal-reacts-with-hydrochloric-acid-according-to-the-following-balanced-equation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/02\/zinc-metal-reacts-with-hydrochloric-acid-according-to-the-following-balanced-equation\/","title":{"rendered":"Zinc metal reacts with hydrochloric acid according to the following balanced equation"},"content":{"rendered":"\n<p>Zinc metal reacts with hydrochloric acid according to the following balanced equation: Zn (s) + 2 HCl (aq) \u00e2\u2020\u2019 ZnCl2 (aq) + H2 (g) When 0.101 g of Zn (s) is combined with enough HCl to make a 50.0 mL solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.27 \u00c2\u00b0C to 23.65 \u00c2\u00b0C. Calculate the enthalpy of reaction (\u00ce\u201dHrxn) in kJ\/mol. Given: Assume that the density of the solution is 1.00 g\/mL and the specific heat of the solution is 4.184 J\/g\u00c2\u00b7\u00c2\u00b0C. IMPORTANT: Enter both the sign and the value (no units) using 3 significant figures<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the enthalpy of reaction (\u0394H\u2093\u2093\u2093) in kJ\/mol, we first need to use the heat released in the reaction, and then determine how much energy was released per mole of zinc reacted.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the heat released (q)<\/h3>\n\n\n\n<p>The formula for heat is:q=m\u00d7c\u00d7\u0394Tq = m \\times c \\times \\Delta Tq=m\u00d7c\u00d7\u0394T<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>mmm = mass of the solution (g)<\/li>\n\n\n\n<li>ccc = specific heat of the solution (J\/g\u00b7\u00b0C)<\/li>\n\n\n\n<li>\u0394T\\Delta T\u0394T = change in temperature (\u00b0C)<\/li>\n<\/ul>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The mass of the solution: Since the density is 1.00 g\/mL and the volume is 50.0 mL, the mass of the solution is equal to the volume in grams: m=50.0\u2009gm = 50.0 \\, \\text{g}m=50.0g<\/li>\n\n\n\n<li>The specific heat of the solution: c=4.184\u2009J\/g\\cdotp\u00b0Cc = 4.184 \\, \\text{J\/g\u00b7\u00b0C}c=4.184J\/g\\cdotp\u00b0C<\/li>\n\n\n\n<li>The temperature change: \u0394T=23.65\u2218C\u221222.27\u2218C=1.38\u2218C\\Delta T = 23.65^\\circ C &#8211; 22.27^\\circ C = 1.38^\\circ C\u0394T=23.65\u2218C\u221222.27\u2218C=1.38\u2218C<\/li>\n<\/ul>\n\n\n\n<p>Now, calculating the heat released:q=50.0\u2009g\u00d74.184\u2009J\/g\\cdotp\u00b0C\u00d71.38\u2218C=288.76\u2009Jq = 50.0 \\, \\text{g} \\times 4.184 \\, \\text{J\/g\u00b7\u00b0C} \\times 1.38^\\circ C = 288.76 \\, \\text{J}q=50.0g\u00d74.184J\/g\\cdotp\u00b0C\u00d71.38\u2218C=288.76J<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the moles of zinc reacted<\/h3>\n\n\n\n<p>The molar mass of zinc (Zn) is 65.38 g\/mol. We are given the mass of zinc used:Mass&nbsp;of&nbsp;Zn=0.101\u2009g\\text{Mass of Zn} = 0.101 \\, \\text{g}Mass&nbsp;of&nbsp;Zn=0.101g<\/p>\n\n\n\n<p>Now, calculate the moles of zinc reacted:Moles&nbsp;of&nbsp;Zn=Mass&nbsp;of&nbsp;ZnMolar&nbsp;mass&nbsp;of&nbsp;Zn=0.101\u2009g65.38\u2009g\/mol=0.001544\u2009mol\\text{Moles of Zn} = \\frac{\\text{Mass of Zn}}{\\text{Molar mass of Zn}} = \\frac{0.101 \\, \\text{g}}{65.38 \\, \\text{g\/mol}} = 0.001544 \\, \\text{mol}Moles&nbsp;of&nbsp;Zn=Molar&nbsp;mass&nbsp;of&nbsp;ZnMass&nbsp;of&nbsp;Zn\u200b=65.38g\/mol0.101g\u200b=0.001544mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the enthalpy change per mole of zinc<\/h3>\n\n\n\n<p>Now that we know the heat released (qqq) and the moles of zinc reacted, we can calculate the enthalpy of reaction per mole of zinc:\u0394Hrxn=qmoles&nbsp;of&nbsp;Zn\\Delta H_{\\text{rxn}} = \\frac{q}{\\text{moles of Zn}}\u0394Hrxn\u200b=moles&nbsp;of&nbsp;Znq\u200b<\/p>\n\n\n\n<p>Substitute the values:\u0394Hrxn=288.76\u2009J0.001544\u2009mol=187,000\u2009J\/mol=187\u2009kJ\/mol\\Delta H_{\\text{rxn}} = \\frac{288.76 \\, \\text{J}}{0.001544 \\, \\text{mol}} = 187,000 \\, \\text{J\/mol} = 187 \\, \\text{kJ\/mol}\u0394Hrxn\u200b=0.001544mol288.76J\u200b=187,000J\/mol=187kJ\/mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>\u0394Hrxn=\u2212187\u2009kJ\/mol\\Delta H_{\\text{rxn}} = -187 \\, \\text{kJ\/mol}\u0394Hrxn\u200b=\u2212187kJ\/mol<\/p>\n\n\n\n<p>The negative sign indicates that the reaction is exothermic (heat is released). So, the enthalpy of reaction is <strong>-187 kJ\/mol<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-34.jpeg\" alt=\"\" class=\"wp-image-239491\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Zinc metal reacts with hydrochloric acid according to the following balanced equation: Zn (s) + 2 HCl (aq) \u00e2\u2020\u2019 ZnCl2 (aq) + H2 (g) When 0.101 g of Zn (s) is combined with enough HCl to make a 50.0 mL solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the [&hellip;]<\/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-239490","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239490","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=239490"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239490\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}