An aqueous solution containing 9.83 g of lead(II) nitrate is added to an aqueous solution containing 6.94 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical equation: Pb(NO3)2(aq) + 2KCl(aq) What is the limiting reactant? potassium chloride lead(II) nitrate PbCl2 (s) + 2KNO3(aq) The percent yield for the reaction is 87.2%. How many grams of the precipitate are formed?<\/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 lead(II) nitrate and potassium chloride is:<\/p>\n\n\n\n [ To determine the limiting reactant, we need to first calculate the number of moles of each reactant.<\/p>\n\n\n\n The molar mass of lead(II) nitrate, Pb(NO\u2083)\u2082, is calculated as:<\/p>\n\n\n\n [ The moles of Pb(NO\u2083)\u2082 are:<\/p>\n\n\n\n [ The molar mass of potassium chloride (KCl) is:<\/p>\n\n\n\n [ The moles of KCl are:<\/p>\n\n\n\n [ From the balanced chemical equation, we know that 1 mole of Pb(NO\u2083)\u2082 reacts with 2 moles of KCl. Therefore, the required moles of KCl to completely react with 0.0297 mol of Pb(NO\u2083)\u2082 are:<\/p>\n\n\n\n [ Since we have 0.0931 mol of KCl, which is more than enough to react with the 0.0297 mol of Pb(NO\u2083)\u2082, lead(II) nitrate (Pb(NO\u2083)\u2082) is the limiting reactant.<\/p>\n\n\n\n From the balanced equation, 1 mole of Pb(NO\u2083)\u2082 produces 1 mole of PbCl\u2082. Therefore, 0.0297 moles of Pb(NO\u2083)\u2082 will produce 0.0297 moles of PbCl\u2082.<\/p>\n\n\n\n The molar mass of PbCl\u2082 is:<\/p>\n\n\n\n [ The theoretical yield of PbCl\u2082 is:<\/p>\n\n\n\n [ The percent yield is 87.2%. Therefore, the actual yield of PbCl\u2082 is:<\/p>\n\n\n\n [ The limiting reactant is lead(II) nitrate (Pb(NO\u2083)\u2082)<\/strong>, and the amount of precipitate formed, considering the 87.2% yield, is 7.22 grams<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" An aqueous solution containing 9.83 g of lead(II) nitrate is added to an aqueous solution containing 6.94 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical equation: Pb(NO3)2(aq) + 2KCl(aq) What is the limiting reactant? potassium chloride lead(II) nitrate PbCl2 (s) + 2KNO3(aq) […]<\/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-189601","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189601","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=189601"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189601\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189601"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189601"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189601"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Balanced Chemical Equation:<\/h3>\n\n\n\n
\\text{Pb(NO}_3\\text{)}_2(aq) + 2\\text{KCl}(aq) \\rightarrow \\text{PbCl}_2(s) + 2\\text{KNO}_3(aq)
]<\/p>\n\n\n\nDetermining the Limiting Reactant:<\/h3>\n\n\n\n
Moles of Lead(II) Nitrate (Pb(NO\u2083)\u2082):<\/h4>\n\n\n\n
\\text{Molar mass of Pb(NO}_3\\text{)}_2 = 207.2 \\, (\\text{Pb}) + 2 \\times (14.01 \\, (\\text{N}) + 3 \\times 16.00 \\, (\\text{O})) = 331.2 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{moles of Pb(NO}_3\\text{)}_2 = \\frac{9.83 \\, \\text{g}}{331.2 \\, \\text{g\/mol}} = 0.0297 \\, \\text{mol}
]<\/p>\n\n\n\nMoles of Potassium Chloride (KCl):<\/h4>\n\n\n\n
\\text{Molar mass of KCl} = 39.10 \\, (\\text{K}) + 35.45 \\, (\\text{Cl}) = 74.55 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{moles of KCl} = \\frac{6.94 \\, \\text{g}}{74.55 \\, \\text{g\/mol}} = 0.0931 \\, \\text{mol}
]<\/p>\n\n\n\nStoichiometric Calculation:<\/h3>\n\n\n\n
\\text{Required moles of KCl} = 2 \\times 0.0297 = 0.0594 \\, \\text{mol}
]<\/p>\n\n\n\nCalculating the Theoretical Yield of PbCl\u2082:<\/h3>\n\n\n\n
\\text{Molar mass of PbCl}_2 = 207.2 \\, (\\text{Pb}) + 2 \\times 35.45 \\, (\\text{Cl}) = 278.1 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{Theoretical yield of PbCl}_2 = 0.0297 \\, \\text{mol} \\times 278.1 \\, \\text{g\/mol} = 8.27 \\, \\text{g}
]<\/p>\n\n\n\nCalculating the Actual Yield:<\/h3>\n\n\n\n
\\text{Actual yield of PbCl}_2 = \\frac{87.2}{100} \\times 8.27 \\, \\text{g} = 7.22 \\, \\text{g}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n