Potassium Chloride Reacts With Silver Nitrate To Produce Solid Silver Chloride And Potassium Nitrate. KCl(Aq) + AgNO3(Aq) \u2192 AgCl(S) + KNO3(Aq) How Many Grams Of KCl Are Required To React Completely With 250 ML Of 0.030 M AgNO3? 2.63 G 13.18 0.1018 3.358 0.1128 0.559 8<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The balanced chemical equation for the reaction between potassium chloride (KCl) and silver nitrate (AgNO\u2083) is:<\/p>\n\n\n\n [ \\text{KCl (aq)} + \\text{AgNO\u2083 (aq)} \\rightarrow \\text{AgCl (s)} + \\text{KNO\u2083 (aq)} ]<\/p>\n\n\n\n First, we need to find the moles of AgNO\u2083 in the solution. The molarity (M) of AgNO\u2083 is 0.030 M, and the volume of the solution is 250 mL, or 0.250 L.<\/p>\n\n\n\n Moles of AgNO\u2083 can be calculated using the formula:<\/p>\n\n\n\n [ [ From the balanced equation, we can see that the mole ratio of KCl to AgNO\u2083 is 1:1. This means that 1 mole of KCl reacts with 1 mole of AgNO\u2083. Therefore, the moles of KCl required to react with 0.0075 moles of AgNO\u2083 is also 0.0075 moles of KCl.<\/p>\n\n\n\n Now, we will convert the moles of KCl to grams. The molar mass of KCl is calculated as:<\/p>\n\n\n\n [ The mass of KCl required is:<\/p>\n\n\n\n [ [ The mass of KCl required to react completely with 250 mL of 0.030 M AgNO\u2083 is 0.559 g<\/strong>.<\/p>\n\n\n\n Thus, the correct answer is 0.559 g<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Potassium Chloride Reacts With Silver Nitrate To Produce Solid Silver Chloride And Potassium Nitrate. KCl(Aq) + AgNO3(Aq) \u2192 AgCl(S) + KNO3(Aq) How Many Grams Of KCl Are Required To React Completely With 250 ML Of 0.030 M AgNO3? 2.63 G 13.18 0.1018 3.358 0.1128 0.559 8 The Correct Answer and Explanation is : The balanced […]<\/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-186117","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186117","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=186117"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186117\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186117"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186117"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186117"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate the moles of AgNO\u2083<\/h3>\n\n\n\n
\\text{Moles of AgNO\u2083} = \\text{Molarity} \\times \\text{Volume}
]<\/p>\n\n\n\n
\\text{Moles of AgNO\u2083} = 0.030 \\, \\text{M} \\times 0.250 \\, \\text{L} = 0.0075 \\, \\text{mol}
]<\/p>\n\n\n\nStep 2: Use the stoichiometry of the reaction<\/h3>\n\n\n\n
Step 3: Calculate the mass of KCl required<\/h3>\n\n\n\n
\\text{Molar mass of KCl} = \\text{K} + \\text{Cl} = 39.1 \\, \\text{g\/mol} + 35.45 \\, \\text{g\/mol} = 74.55 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{Mass of KCl} = \\text{Moles of KCl} \\times \\text{Molar mass of KCl}
]<\/p>\n\n\n\n
\\text{Mass of KCl} = 0.0075 \\, \\text{mol} \\times 74.55 \\, \\text{g\/mol} = 0.559 \\, \\text{g}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n