{"id":38607,"date":"2025-06-26T10:55:56","date_gmt":"2025-06-26T10:55:56","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38607"},"modified":"2025-06-26T10:55:57","modified_gmt":"2025-06-26T10:55:57","slug":"how-many-grams-of-nh3-can-be-produced-from-2-55-mol-of-n2-and-excess-h2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-grams-of-nh3-can-be-produced-from-2-55-mol-of-n2-and-excess-h2\/","title":{"rendered":"How many grams of NH3 can be produced from 2.55 mol of N2 and excess H2"},"content":{"rendered":"\n<p>How many grams of NH3 can be produced from 2.55 mol of N2 and excess H2? How many grams of H2 are needed to produce 10.38 g of NH3? How many molecules (not moles) of NH3 are produced from 7.87&#215;10^-4 g of H2?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To answer the questions, we need to use the balanced chemical equation for the synthesis of ammonia (NH\u2083):<\/p>\n\n\n\n<p><strong>N\u2082 + 3H\u2082 \u2192 2NH\u2083<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. How many grams of NH\u2083 can be produced from 2.55 mol of N\u2082 and excess H\u2082?<\/strong><\/h3>\n\n\n\n<p>From the balanced equation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mol of N\u2082 produces 2 mol of NH\u2083<\/li>\n\n\n\n<li>So, 2.55 mol of N\u2082 will produce:<br>2.55 mol N\u2082 \u00d7 (2 mol NH\u2083 \/ 1 mol N\u2082) = <strong>5.10 mol NH\u2083<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Now convert moles of NH\u2083 to grams:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of NH\u2083 = 14.01 (N) + 3 \u00d7 1.008 (H) = 17.034 g\/mol<\/li>\n\n\n\n<li>Mass = 5.10 mol \u00d7 17.034 g\/mol = <strong>86.87 g NH\u2083<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. How many grams of H\u2082 are needed to produce 10.38 g of NH\u2083?<\/strong><\/h3>\n\n\n\n<p>First, convert grams of NH\u2083 to moles:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10.38 g \u00f7 17.034 g\/mol = <strong>0.6096 mol NH\u2083<\/strong><\/li>\n<\/ul>\n\n\n\n<p>From the balanced equation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 mol NH\u2083 are produced from 3 mol H\u2082<\/li>\n\n\n\n<li>0.6096 mol NH\u2083 \u00d7 (3 mol H\u2082 \/ 2 mol NH\u2083) = <strong>0.9144 mol H\u2082<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Convert moles of H\u2082 to grams:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of H\u2082 = 2.016 g\/mol<\/li>\n\n\n\n<li>0.9144 mol \u00d7 2.016 g\/mol = <strong>1.844 g H\u2082<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>3. How many molecules of NH\u2083 are produced from 7.87\u00d710\u207b\u2074 g of H\u2082?<\/strong><\/h3>\n\n\n\n<p>Convert grams of H\u2082 to moles:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>7.87\u00d710\u207b\u2074 g \u00f7 2.016 g\/mol = <strong>3.905\u00d710\u207b\u2074 mol H\u2082<\/strong><\/li>\n<\/ul>\n\n\n\n<p>From the equation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 mol H\u2082 produce 2 mol NH\u2083<\/li>\n\n\n\n<li>3.905\u00d710\u207b\u2074 mol H\u2082 \u00d7 (2 mol NH\u2083 \/ 3 mol H\u2082) = <strong>2.603\u00d710\u207b\u2074 mol NH\u2083<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Convert moles to molecules:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Use Avogadro&#8217;s number: 6.022\u00d710\u00b2\u00b3 molecules\/mol<\/li>\n\n\n\n<li>2.603\u00d710\u207b\u2074 mol \u00d7 6.022\u00d710\u00b2\u00b3 = <strong>1.568\u00d710\u00b2\u2070 molecules of NH\u2083<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>To solve chemical problems like these, it is essential to begin with a balanced chemical equation. The synthesis of ammonia, a common reaction in industrial chemistry, follows this formula: <strong>N\u2082 + 3H\u2082 \u2192 2NH\u2083<\/strong>. This shows that one molecule of nitrogen reacts with three molecules of hydrogen to form two molecules of ammonia. The stoichiometric coefficients (1, 3, and 2) help determine the mole ratios needed for calculations.<\/p>\n\n\n\n<p>In the first part, we are given 2.55 mol of nitrogen gas and asked to find how much ammonia is produced, assuming hydrogen is in excess. Since one mole of nitrogen produces two moles of ammonia, multiplying 2.55 mol by 2 gives 5.10 mol NH\u2083. Using ammonia\u2019s molar mass, we convert moles to grams.<\/p>\n\n\n\n<p>The second question starts with a mass of ammonia, and we work backward. First, we convert that mass to moles using the molar mass of ammonia. Then, we use the mole ratio to find how many moles of hydrogen gas are required. Finally, converting those moles to grams gives the mass of hydrogen needed.<\/p>\n\n\n\n<p>In the final question, we are given a tiny mass of hydrogen gas and asked how many ammonia molecules it can produce. After converting grams of hydrogen to moles, we again apply the mole ratio to find the number of moles of ammonia produced. We then multiply by Avogadro\u2019s number to get the exact number of ammonia molecules.<\/p>\n\n\n\n<p>These problems demonstrate the importance of understanding mole ratios, molar masses, and the conversion between moles and molecules in chemistry.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1043.jpeg\" alt=\"\" class=\"wp-image-38608\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1043.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1043-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1043-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>How many grams of NH3 can be produced from 2.55 mol of N2 and excess H2? How many grams of H2 are needed to produce 10.38 g of NH3? How many molecules (not moles) of NH3 are produced from 7.87&#215;10^-4 g of H2? The Correct Answer and Explanation is: To answer the questions, we need [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-38607","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38607","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=38607"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38607\/revisions"}],"predecessor-version":[{"id":38609,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38607\/revisions\/38609"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38607"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38607"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38607"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}