{"id":209369,"date":"2025-04-29T05:02:47","date_gmt":"2025-04-29T05:02:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=209369"},"modified":"2025-04-29T05:02:49","modified_gmt":"2025-04-29T05:02:49","slug":"the-molar-mass-of-nh3-is-17-03-g-mol-and-the-molar-mass-of-h2o-is-18-02-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/29\/the-molar-mass-of-nh3-is-17-03-g-mol-and-the-molar-mass-of-h2o-is-18-02-g-mol\/","title":{"rendered":"The molar mass of NH3 is 17.03 g\/mol, and the molar mass of H2O is 18.02 g\/mol"},"content":{"rendered":"\n<p>The molar mass of NH3 is 17.03 g\/mol, and the molar mass of H2O is 18.02 g\/mol. Consider the following (already balanced) chemical equation: 4 NH3(g) + 7 O2(g) ? 4 NO2(g) + 6 H2O(g) How many moles of H2O are formed when 58.6 g of NH3 react with excess oxygen according to the above equation?<\/p>\n\n\n\n<p>1.50 x 10<br>2.29<br>3.44<br>87.9<br>5.16<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s solve it carefully first:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of NH\u2083 = 17.03 g\/mol<\/li>\n\n\n\n<li>Balanced reaction:<br><strong>4 NH\u2083(g) + 7 O\u2082(g) \u2192 4 NO\u2082(g) + 6 H\u2082O(g)<\/strong><\/li>\n\n\n\n<li>Mass of NH\u2083 = 58.6 g<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 1: Find moles of NH\u2083.<\/strong><br>[<br>\\text{moles of NH}_3 = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{58.6\\ \\text{g}}{17.03\\ \\text{g\/mol}} = 3.44\\ \\text{mol}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 2: Use the mole ratio from the balanced equation.<\/strong><br>From the reaction,<br>4 moles of NH\u2083 produce 6 moles of H\u2082O.<\/p>\n\n\n\n<p>Set up the proportion:<br>[<br>\\text{moles of H}_2\\text{O} = 3.44\\ \\text{mol NH}_3 \\times \\frac{6\\ \\text{mol H}_2\\text{O}}{4\\ \\text{mol NH}_3}<br>]<br>[<br>= 3.44 \\times 1.5<br>]<br>[<br>= 5.16\\ \\text{mol H}_2\\text{O}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>\u2705 <strong>Final Answer: 5.16<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\">Explanation (Approx. 300 words):<\/h1>\n\n\n\n<p>In this question, we are asked to find out how many moles of water (H\u2082O) are produced when a known mass of ammonia (NH\u2083) reacts completely with excess oxygen. The first step is to convert the mass of NH\u2083 into moles, because chemical equations work in terms of moles, not grams.<\/p>\n\n\n\n<p>The molar mass of NH\u2083 is given as 17.03 g\/mol. Using the basic formula for moles, which is mass divided by molar mass, we find that:<br>[<br>\\text{moles of NH}_3 = \\frac{58.6\\ \\text{g}}{17.03\\ \\text{g\/mol}} = 3.44\\ \\text{mol}<br>]<\/p>\n\n\n\n<p>Next, we refer to the balanced chemical equation:<br>[<br>4\\ \\text{NH}_3 + 7\\ \\text{O}_2 \\rightarrow 4\\ \\text{NO}_2 + 6\\ \\text{H}_2\\text{O}<br>]<br>This equation tells us that <strong>4 moles of NH\u2083<\/strong> yield <strong>6 moles of H\u2082O<\/strong>. Thus, for every mole of NH\u2083, 1.5 moles of H\u2082O are formed.<\/p>\n\n\n\n<p>Multiplying the moles of NH\u2083 (3.44 mol) by the ratio (1.5), we get:<br>[<br>3.44\\ \\text{mol} \\times 1.5 = 5.16\\ \\text{mol H}_2\\text{O}<br>]<\/p>\n\n\n\n<p>This calculation assumes that oxygen is in excess, so NH\u2083 is the limiting reactant.<\/p>\n\n\n\n<p>Finally, we conclude that 5.16 moles of water are produced when 58.6 grams of ammonia react under these conditions.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n","protected":false},"excerpt":{"rendered":"<p>The molar mass of NH3 is 17.03 g\/mol, and the molar mass of H2O is 18.02 g\/mol. Consider the following (already balanced) chemical equation: 4 NH3(g) + 7 O2(g) ? 4 NO2(g) + 6 H2O(g) How many moles of H2O are formed when 58.6 g of NH3 react with excess oxygen according to the above [&hellip;]<\/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-209369","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209369","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=209369"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209369\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209369"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209369"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209369"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}