When solutions of silver nitrate and potassium chloride are mixed, silver chloride precipitates out of solution according to the equation AgNO3(aq)+KCl(aq)\u2192AgCl(s) KNO3(aq) Part A What mass of silver chloride can be produced from 1.88 L of a 0.160 M solution of silver nitrate? Express your answer with the appropriate units<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the mass of silver chloride (AgCl) produced when silver nitrate (AgNO\u2083) reacts with potassium chloride (KCl), we need to go through several steps involving stoichiometry, starting from the balanced chemical equation:<\/p>\n\n\n\n [ This reaction shows a 1:1 molar ratio between silver nitrate and silver chloride. Thus, one mole of silver nitrate produces one mole of silver chloride.<\/p>\n\n\n\n The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. We are given:<\/p>\n\n\n\n The number of moles of AgNO\u2083 can be calculated using the formula: From the balanced equation, we know that for every mole of AgNO\u2083, one mole of AgCl is produced. Therefore, the moles of AgCl produced will also be 0.3008 mol.<\/p>\n\n\n\n Now, to find the mass of silver chloride, we use the molar mass of AgCl. The molar mass of AgCl is calculated as:<\/p>\n\n\n\n Thus, the molar mass of AgCl is: Now, using the formula: The mass of silver chloride (AgCl) that can be produced from 1.88 L of a 0.160 M solution of silver nitrate is 43.12 grams<\/strong>.<\/p>\n\n\n\n In this process, we used stoichiometry to convert the volume and concentration of the silver nitrate solution into moles. The mole-to-mole ratio from the balanced equation tells us that the moles of AgNO\u2083 will be equivalent to the moles of AgCl produced. By multiplying the moles of AgCl by its molar mass, we determined the mass of the silver chloride that precipitates from the solution.<\/p>\n","protected":false},"excerpt":{"rendered":" When solutions of silver nitrate and potassium chloride are mixed, silver chloride precipitates out of solution according to the equation AgNO3(aq)+KCl(aq)\u2192AgCl(s) KNO3(aq) Part A What mass of silver chloride can be produced from 1.88 L of a 0.160 M solution of silver nitrate? Express your answer with the appropriate units The Correct Answer and Explanation […]<\/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-186119","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186119","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=186119"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186119\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186119"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186119"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{AgNO\u2083 (aq)} + \\text{KCl (aq)} \\rightarrow \\text{AgCl (s)} + \\text{KNO\u2083 (aq)}
]<\/p>\n\n\n\nStep 1: Calculate the moles of AgNO\u2083<\/h3>\n\n\n\n
\n
[
\\text{moles of AgNO\u2083} = M \\times V
]
[
\\text{moles of AgNO\u2083} = 0.160 \\, \\text{mol\/L} \\times 1.88 \\, \\text{L} = 0.3008 \\, \\text{mol}
]<\/p>\n\n\n\nStep 2: Determine the moles of AgCl produced<\/h3>\n\n\n\n
Step 3: Calculate the mass of AgCl<\/h3>\n\n\n\n
\n
[
\\text{Molar mass of AgCl} = 107.87 + 35.45 = 143.32 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
[
\\text{mass of AgCl} = \\text{moles of AgCl} \\times \\text{molar mass of AgCl}
]
[
\\text{mass of AgCl} = 0.3008 \\, \\text{mol} \\times 143.32 \\, \\text{g\/mol} = 43.12 \\, \\text{g}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n