Let’s break down the reaction step by step, starting with the balanced equation:<\/p>\n\n\n\n
Balanced equation:<\/strong> We are given 25 grams of lead (II) nitrate (Pb(NO\u2083)\u2082) and 15.0 grams of sodium iodide (NaI). To find the limiting reagent, we first need to convert the masses into moles.<\/p>\n\n\n\n Molar Mass of Pb(NO\u2083)\u2082:<\/strong> Moles of Pb(NO\u2083)\u2082:<\/strong> Moles of Pb(NO\u2083)\u2082=25.0\u2009g241.2\u2009g\/mol=0.1036\u2009mol\\text{Moles of Pb(NO\u2083)\u2082} = \\frac{25.0 \\, \\text{g}}{241.2 \\, \\text{g\/mol}} = 0.1036 \\, \\text{mol}Moles of Pb(NO\u2083)\u2082=241.2g\/mol25.0g\u200b=0.1036mol<\/p>\n\n\n\n Molar Mass of NaI:<\/strong> Na=22.99\u2009g\/mol,\u2009I=126.9\u2009g\/mol\\text{Na} = 22.99 \\, \\text{g\/mol}, \\, \\text{I} = 126.9 \\, \\text{g\/mol}Na=22.99g\/mol,I=126.9g\/mol Molar Mass of NaI=22.99+126.9=149.89\u2009g\/mol\\text{Molar Mass of NaI} = 22.99 + 126.9 = 149.89 \\, \\text{g\/mol}Molar Mass of NaI=22.99+126.9=149.89g\/mol<\/p>\n\n\n\n Moles of NaI:<\/strong> Moles of NaI=15.0\u2009g149.89\u2009g\/mol=0.1001\u2009mol\\text{Moles of NaI} = \\frac{15.0 \\, \\text{g}}{149.89 \\, \\text{g\/mol}} = 0.1001 \\, \\text{mol}Moles of NaI=149.89g\/mol15.0g\u200b=0.1001mol<\/p>\n\n\n\n From the balanced equation, 1 mole of Pb(NO\u2083)\u2082 reacts with 2 moles of NaI. So, for 0.1036 moles of Pb(NO\u2083)\u2082, you would need: 0.1036\u2009mol Pb(NO\u2083)\u2082\u00d72\u2009mol NaI1\u2009mol Pb(NO\u2083)\u2082=0.2072\u2009mol NaI0.1036 \\, \\text{mol Pb(NO\u2083)\u2082} \\times \\frac{2 \\, \\text{mol NaI}}{1 \\, \\text{mol Pb(NO\u2083)\u2082}} = 0.2072 \\, \\text{mol NaI}0.1036mol Pb(NO\u2083)\u2082\u00d71mol Pb(NO\u2083)\u20822mol NaI\u200b=0.2072mol NaI<\/p>\n\n\n\n However, we only have 0.1001 moles of NaI, which is less than the required 0.2072 moles. Therefore, NaI is the limiting reagent<\/strong>.<\/p>\n\n\n\n From the balanced equation, 2 moles of NaI produce 2 moles of NaNO\u2083. Since NaI is the limiting reagent, we can directly calculate the moles of NaNO\u2083 produced: Moles of NaNO\u2083=0.1001\u2009mol NaI\u00d72\u2009mol NaNO\u20832\u2009mol NaI=0.1001\u2009mol NaNO\u2083\\text{Moles of NaNO\u2083} = 0.1001 \\, \\text{mol NaI} \\times \\frac{2 \\, \\text{mol NaNO\u2083}}{2 \\, \\text{mol NaI}} = 0.1001 \\, \\text{mol NaNO\u2083}Moles of NaNO\u2083=0.1001mol NaI\u00d72mol NaI2mol NaNO\u2083\u200b=0.1001mol NaNO\u2083<\/p>\n\n\n\n Molar Mass of NaNO\u2083:<\/strong> NaNO\u2083=22.99+14.0+3\u00d716.0=85.0\u2009g\/mol\\text{NaNO\u2083} = 22.99 + 14.0 + 3 \\times 16.0 = 85.0 \\, \\text{g\/mol}NaNO\u2083=22.99+14.0+3\u00d716.0=85.0g\/mol<\/p>\n\n\n\n Mass of NaNO\u2083 formed:<\/strong> Mass of NaNO\u2083=0.1001\u2009mol\u00d785.0\u2009g\/mol=8.51\u2009g\\text{Mass of NaNO\u2083} = 0.1001 \\, \\text{mol} \\times 85.0 \\, \\text{g\/mol} = 8.51 \\, \\text{g}Mass of NaNO\u2083=0.1001mol\u00d785.0g\/mol=8.51g<\/p>\n\n\n\n From the balanced equation, 2 moles of NaI produce 1 mole of PbI\u2082. So, the moles of PbI\u2082 formed are: Moles of PbI\u2082=0.1001\u2009mol NaI\u00d71\u2009mol PbI\u20822\u2009mol NaI=0.05005\u2009mol PbI\u2082\\text{Moles of PbI\u2082} = 0.1001 \\, \\text{mol NaI} \\times \\frac{1 \\, \\text{mol PbI\u2082}}{2 \\, \\text{mol NaI}} = 0.05005 \\, \\text{mol PbI\u2082}Moles of PbI\u2082=0.1001mol NaI\u00d72mol NaI1mol PbI\u2082\u200b=0.05005mol PbI\u2082<\/p>\n\n\n\n Molar Mass of PbI\u2082:<\/strong> PbI\u2082=207.2+2\u00d7126.9=461.0\u2009g\/mol\\text{PbI\u2082} = 207.2 + 2 \\times 126.9 = 461.0 \\, \\text{g\/mol}PbI\u2082=207.2+2\u00d7126.9=461.0g\/mol<\/p>\n\n\n\n Mass of PbI\u2082 formed:<\/strong> Mass of PbI\u2082=0.05005\u2009mol\u00d7461.0\u2009g\/mol=23.1\u2009g\\text{Mass of PbI\u2082} = 0.05005 \\, \\text{mol} \\times 461.0 \\, \\text{g\/mol} = 23.1 \\, \\text{g}Mass of PbI\u2082=0.05005mol\u00d7461.0g\/mol=23.1g<\/p>\n\n\n\n Since NaI is the limiting reagent, there will be no NaI left over because all of it is used up in the reaction.<\/p>\n\n\n\n We are told that 6 grams of NaNO\u2083 are formed, but the theoretical yield was calculated as 8.51 grams.<\/p>\n\n\n\n Percent Yield:<\/strong> Percent Yield=Actual YieldTheoretical Yield\u00d7100=6.0\u2009g8.51\u2009g\u00d7100=70.5%\\text{Percent Yield} = \\frac{\\text{Actual Yield}}{\\text{Theoretical Yield}} \\times 100 = \\frac{6.0 \\, \\text{g}}{8.51 \\, \\text{g}} \\times 100 = 70.5\\%Percent Yield=Theoretical YieldActual Yield\u200b\u00d7100=8.51g6.0g\u200b\u00d7100=70.5%<\/p>\n\n\n\n When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed. Balance the following equation: Pb(NO3)2 (aq) + 2NaI (aq) \u00e2\u2020\u2019 2NaNO3 (aq) + PbI2 (s) If I start with 25 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be […]<\/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-267942","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267942","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=267942"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267942\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=267942"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=267942"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=267942"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Pb(NO3)2(aq)+2NaI(aq)\u21922NaNO3(aq)+PbI2(s)\\text{Pb(NO}_3)_2 (aq) + 2\\text{NaI} (aq) \\rightarrow 2\\text{NaNO}_3 (aq) + \\text{PbI}_2 (s)Pb(NO3\u200b)2\u200b(aq)+2NaI(aq)\u21922NaNO3\u200b(aq)+PbI2\u200b(s)<\/p>\n\n\n\n1. Identifying the Limiting Reagent:<\/strong><\/h3>\n\n\n\n
Pb=207.2\u2009g\/mol,\u2009N=14.0\u2009g\/mol,\u2009O=16.0\u2009g\/mol\\text{Pb} = 207.2 \\, \\text{g\/mol}, \\, \\text{N} = 14.0 \\, \\text{g\/mol}, \\, \\text{O} = 16.0 \\, \\text{g\/mol}Pb=207.2g\/mol,N=14.0g\/mol,O=16.0g\/mol
Molar Mass of Pb(NO\u2083)\u2082=207.2+2(14.0+3\u00d716.0)=241.2\u2009g\/mol\\text{Molar Mass of Pb(NO\u2083)\u2082} = 207.2 + 2(14.0 + 3 \\times 16.0) = 241.2 \\, \\text{g\/mol}Molar Mass of Pb(NO\u2083)\u2082=207.2+2(14.0+3\u00d716.0)=241.2g\/mol<\/p>\n\n\n\n2. Stoichiometry of the Reaction:<\/strong><\/h3>\n\n\n\n
3. Amount of Sodium Nitrate Formed:<\/strong><\/h3>\n\n\n\n
4. Amount of Lead (II) Iodide Formed:<\/strong><\/h3>\n\n\n\n
5. Excess NaI Left Over:<\/strong><\/h3>\n\n\n\n
6. Percent Yield:<\/strong><\/h3>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
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