Calcium carbonate reacts with hydrochloric acid (HCl(aq)) to form calcium chloride and carbon dioxide and water.<\/p>\n\n\n\n
a) How much calcium chloride would be formed if 14.7g of calcium carbonate reacted with 250 mL of a 1.6 M solution of hydrochloric acid (HCl(aq)?<\/p>\n\n\n\n
b) Calculate the percent yield if 10.6g of calcium chloride was produced.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we need to follow a systematic approach:<\/p>\n\n\n\n a) How much calcium chloride would be formed if 14.7g of calcium carbonate reacted with 250 mL of a 1.6 M solution of hydrochloric acid (HCl(aq))?<\/strong><\/p>\n\n\n\n b) Calculate the percent yield if 10.6g of calcium chloride was produced.<\/strong><\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n In this reaction, calcium carbonate reacts with hydrochloric acid to produce calcium chloride, carbon dioxide, and water. To determine how much calcium chloride is formed, we first need to identify the limiting reagent. By calculating the moles of each reactant, we find that calcium carbonate is the limiting reagent, so the maximum amount of calcium chloride produced is based on the moles of calcium carbonate. Using stoichiometry, we can convert moles of CaCO\u2083 to moles of CaCl\u2082 and then calculate the theoretical mass of CaCl\u2082.<\/p>\n\n\n\n Next, the percent yield is determined by comparing the actual yield (the amount of CaCl\u2082 actually produced, 10.6g) to the theoretical yield (the maximum amount that could be produced, 16.3g). A yield of 65.0% indicates that not all of the calcium carbonate reacted efficiently, which could be due to side reactions, incomplete reactions, or losses during handling.<\/p>\n","protected":false},"excerpt":{"rendered":" Calcium carbonate reacts with hydrochloric acid (HCl(aq)) to form calcium chloride and carbon dioxide and water. a) How much calcium chloride would be formed if 14.7g of calcium carbonate reacted with 250 mL of a 1.6 M solution of hydrochloric acid (HCl(aq)? b) Calculate the percent yield if 10.6g of calcium chloride was produced. The […]<\/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-182109","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182109","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=182109"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182109\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182109"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182109"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182109"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
[
\\text{CaCO}_3 (s) + 2 \\text{HCl} (aq) \\rightarrow \\text{CaCl}_2 (aq) + \\text{CO}_2 (g) + \\text{H}_2\\text{O} (l)
]<\/li>\n\n\n\n
[
\\text{Molar mass of CaCO}_3 = 40.08 \\, (\\text{Ca}) + 12.00 \\, (\\text{C}) + 3 \\times 16.00 \\, (\\text{O}) = 100.08 \\, \\text{g\/mol}
]
[
\\text{Moles of CaCO}_3 = \\frac{14.7 \\, \\text{g}}{100.08 \\, \\text{g\/mol}} = 0.147 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{volume of solution in liters}}
]
[
\\text{Moles of HCl} = 1.6 \\, \\text{M} \\times 0.250 \\, \\text{L} = 0.400 \\, \\text{mol}
]<\/li>\n\n\n\n
From the balanced equation, the mole ratio of CaCO\u2083 to HCl is 1:2. So, for 0.147 mol of CaCO\u2083, we need:
[
0.147 \\, \\text{mol} \\times 2 = 0.294 \\, \\text{mol of HCl}
]
Since we have 0.400 mol of HCl, HCl is in excess, and CaCO\u2083 is the limiting reagent.<\/li>\n\n\n\n
From the balanced equation, 1 mol of CaCO\u2083 produces 1 mol of CaCl\u2082. So, the moles of CaCl\u2082 formed are 0.147 mol.<\/li>\n\n\n\n
The molar mass of CaCl\u2082 is:
[
\\text{Molar mass of CaCl}_2 = 40.08 \\, (\\text{Ca}) + 2 \\times 35.45 \\, (\\text{Cl}) = 110.98 \\, \\text{g\/mol}
]
[
\\text{Mass of CaCl}_2 = 0.147 \\, \\text{mol} \\times 110.98 \\, \\text{g\/mol} = 16.3 \\, \\text{g}
]<\/li>\n<\/ol>\n\n\n\n\n
[
\\text{Percent yield} = \\frac{\\text{Actual yield}}{\\text{Theoretical yield}} \\times 100
]
[
\\text{Percent yield} = \\frac{10.6 \\, \\text{g}}{16.3 \\, \\text{g}} \\times 100 = 65.0\\%
]<\/li>\n<\/ol>\n\n\n\n