ChatGPT said:<\/h6>\n\n\n\n
To determine the percent yield of sodium nitrate in the reaction between lead (II) nitrate and sodium iodide, let’s break down the steps and calculations:<\/p>\n\n\n\n
Step 1: Balanced Equation<\/h3>\n\n\n\n
The balanced chemical equation for the reaction is:
Pb(NO3)2(aq)+2NaI(aq)\u2192PbI2(s)+2NaNO3(aq)\\text{Pb(NO}_3)_2 (aq) + 2 \\text{NaI} (aq) \\rightarrow \\text{PbI}_2 (s) + 2 \\text{NaNO}_3 (aq)Pb(NO3\u200b)2\u200b(aq)+2NaI(aq)\u2192PbI2\u200b(s)+2NaNO3\u200b(aq)<\/p>\n\n\n\n
Step 2: Moles of Reactants<\/h3>\n\n\n\n
We start by calculating the number of moles of each reactant:<\/p>\n\n\n\n
- \n
- For lead (II) nitrate (Pb(NO\u2083)\u2082):<\/strong>\n
- \n
- Molar mass of Pb(NO\u2083)\u2082 = 331.2 g\/mol<\/li>\n\n\n\n
- Moles of Pb(NO\u2083)\u2082 = 25.0\u00a0g331.2\u00a0g\/mol=0.0755\u00a0mol\\frac{25.0 \\text{ g}}{331.2 \\text{ g\/mol}} = 0.0755 \\text{ mol}331.2\u00a0g\/mol25.0\u00a0g\u200b=0.0755\u00a0mol<\/li>\n<\/ul>\n<\/li>\n\n\n\n
- For sodium iodide (NaI):<\/strong>\n
- \n
- Molar mass of NaI = 149.9 g\/mol<\/li>\n\n\n\n
- Moles of NaI = 15.0\u00a0g149.9\u00a0g\/mol=0.1001\u00a0mol\\frac{15.0 \\text{ g}}{149.9 \\text{ g\/mol}} = 0.1001 \\text{ mol}149.9\u00a0g\/mol15.0\u00a0g\u200b=0.1001\u00a0mol<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
Step 3: Limiting Reagent<\/h3>\n\n\n\n
From the balanced equation, the molar ratio of Pb(NO\u2083)\u2082 to NaI is 1:2. This means:<\/p>\n\n\n\n
- \n
- For every 1 mole of Pb(NO\u2083)\u2082, we need 2 moles of NaI.<\/li>\n\n\n\n
- The number of moles of NaI required to fully react with 0.0755 mol of Pb(NO\u2083)\u2082 is:
0.0755\u00a0mol\u00a0Pb(NO\u2083)\u2082\u00d72=0.151\u00a0mol\u00a0NaI0.0755 \\text{ mol Pb(NO\u2083)\u2082} \\times 2 = 0.151 \\text{ mol NaI}0.0755\u00a0mol\u00a0Pb(NO\u2083)\u2082\u00d72=0.151\u00a0mol\u00a0NaI<\/li>\n<\/ul>\n\n\n\nSince we only have 0.1001 mol of NaI (less than 0.151 mol required), NaI is in excess<\/strong>, and Pb(NO\u2083)\u2082 is the limiting reagent<\/strong>.<\/p>\n\n\n\n
Step 4: Theoretical Yield of Sodium Nitrate<\/h3>\n\n\n\n
According to the balanced equation:<\/p>\n\n\n\n
- \n
- 1 mole of Pb(NO\u2083)\u2082 produces 2 moles of NaNO\u2083.<\/li>\n\n\n\n
- Moles of NaNO\u2083 produced from 0.0755 mol of Pb(NO\u2083)\u2082 = 0.0755\u00a0mol\u00a0Pb(NO\u2083)\u2082\u00d72=0.151\u00a0mol\u00a0NaNO\u20830.0755 \\text{ mol Pb(NO\u2083)\u2082} \\times 2 = 0.151 \\text{ mol NaNO\u2083}0.0755\u00a0mol\u00a0Pb(NO\u2083)\u2082\u00d72=0.151\u00a0mol\u00a0NaNO\u2083<\/li>\n<\/ul>\n\n\n\n
The molar mass of NaNO\u2083 is 85.0 g\/mol, so the theoretical yield of NaNO\u2083 is:<\/p>\n\n\n\n
- \n
- 0.151\u00a0mol\u00a0NaNO\u2083\u00d785.0\u00a0g\/mol=12.9\u00a0g\u00a0NaNO\u20830.151 \\text{ mol NaNO\u2083} \\times 85.0 \\text{ g\/mol} = 12.9 \\text{ g NaNO\u2083}0.151\u00a0mol\u00a0NaNO\u2083\u00d785.0\u00a0g\/mol=12.9\u00a0g\u00a0NaNO\u2083<\/li>\n<\/ul>\n\n\n\n
Step 5: Actual Yield and Percent Yield<\/h3>\n\n\n\n
The actual yield of sodium nitrate is given as 6 grams. The percent yield can be calculated using the formula:
Percent Yield=(Actual YieldTheoretical Yield)\u00d7100\\text{Percent Yield} = \\left( \\frac{\\text{Actual Yield}}{\\text{Theoretical Yield}} \\right) \\times 100Percent Yield=(Theoretical YieldActual Yield\u200b)\u00d7100<\/p>\n\n\n\nSubstituting the values:
Percent Yield=(6 g12.9 g)\u00d7100=46.5%\\text{Percent Yield} = \\left( \\frac{6 \\text{ g}}{12.9 \\text{ g}} \\right) \\times 100 = 46.5\\%Percent Yield=(12.9 g6 g\u200b)\u00d7100=46.5%<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
The percent yield of sodium nitrate in this reaction is 46.5%. This means that only 46.5% of the theoretical amount of sodium nitrate was actually produced, which could be due to experimental losses, impurities, or incomplete reactions.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1727.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead iodide are formed. The balanced chemical equation is: Pb(NO3)2 (aq) + 2 NaI (aq) -> PbI2 (s) + 2 NaNO3 (aq) If we start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, we can calculate how many grams […]<\/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-267951","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267951","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=267951"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267951\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=267951"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=267951"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=267951"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- 0.151\u00a0mol\u00a0NaNO\u2083\u00d785.0\u00a0g\/mol=12.9\u00a0g\u00a0NaNO\u20830.151 \\text{ mol NaNO\u2083} \\times 85.0 \\text{ g\/mol} = 12.9 \\text{ g NaNO\u2083}0.151\u00a0mol\u00a0NaNO\u2083\u00d785.0\u00a0g\/mol=12.9\u00a0g\u00a0NaNO\u2083<\/li>\n<\/ul>\n\n\n\n