{"id":38618,"date":"2025-06-26T11:04:37","date_gmt":"2025-06-26T11:04:37","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38618"},"modified":"2025-06-26T11:04:38","modified_gmt":"2025-06-26T11:04:38","slug":"how-many-grams-of-nh3-can-be-produced-from-the-mixture-of-3-00-g-each-of-nitrogen-and-hydrogen-by-the-haber-process","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-grams-of-nh3-can-be-produced-from-the-mixture-of-3-00-g-each-of-nitrogen-and-hydrogen-by-the-haber-process\/","title":{"rendered":"How many grams of NH3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process"},"content":{"rendered":"\n<p>How many grams of NH3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process? a) 2.00 g b) 3.00 g c) 3.65g d) 1.82 g<\/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 determine how many grams of ammonia (NH\u2083) can be produced from a mixture of <strong>3.00 g of nitrogen (N\u2082)<\/strong> and <strong>3.00 g of hydrogen (H\u2082)<\/strong> using the <strong>Haber process<\/strong>, we begin with the balanced chemical equation:<\/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\">Step 1: Calculate molar masses<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of N\u2082 = 28.02 g\/mol<\/li>\n\n\n\n<li>Molar mass of H\u2082 = 2.02 g\/mol<\/li>\n\n\n\n<li>Molar mass of NH\u2083 = 17.03 g\/mol<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert mass to moles<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of N\u2082 = 3.00 g \u00f7 28.02 g\/mol \u2248 0.107 mol<\/li>\n\n\n\n<li>Moles of H\u2082 = 3.00 g \u00f7 2.02 g\/mol \u2248 1.485 mol<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine limiting reagent<\/h3>\n\n\n\n<p>From the balanced equation, 1 mol of N\u2082 reacts with 3 mol of H\u2082.<br>So, 0.107 mol of N\u2082 would need:<\/p>\n\n\n\n<p><strong>0.107 mol \u00d7 3 = 0.321 mol of H\u2082<\/strong><\/p>\n\n\n\n<p>We have 1.485 mol of H\u2082 available, which is more than enough.<br>Therefore, <strong>N\u2082 is the limiting reagent<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Use limiting reagent to find moles of NH\u2083<\/h3>\n\n\n\n<p>From the equation, 1 mol of N\u2082 produces 2 mol of NH\u2083.<br>So, 0.107 mol of N\u2082 will produce:<\/p>\n\n\n\n<p><strong>0.107 mol \u00d7 2 = 0.214 mol of NH\u2083<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Convert moles of NH\u2083 to grams<\/h3>\n\n\n\n<p><strong>0.214 mol \u00d7 17.03 g\/mol \u2248 3.64 g<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The maximum mass of NH\u2083 that can be produced is approximately <strong>3.65 g<\/strong>, so the correct option is:<\/p>\n\n\n\n<p><strong>c) 3.65 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>This problem demonstrates a classic <strong>limiting reagent<\/strong> scenario in stoichiometry. Even though equal masses of nitrogen and hydrogen are mixed, their different molar masses mean they contribute different numbers of moles. Since nitrogen molecules are heavier, fewer moles of nitrogen are present, and it becomes the limiting reagent. The reaction stops when nitrogen is used up, and the excess hydrogen remains unreacted. The amount of ammonia formed depends entirely on how much nitrogen reacts. By converting grams to moles, applying the balanced equation, and converting the product back to grams, we find the answer.<\/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-banner8-1046.jpeg\" alt=\"\" class=\"wp-image-38619\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1046.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1046-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1046-768x923.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 the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process? a) 2.00 g b) 3.00 g c) 3.65g d) 1.82 g The Correct Answer and Explanation is: To determine how many grams of ammonia (NH\u2083) can be produced from a mixture of [&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-38618","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38618","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=38618"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38618\/revisions"}],"predecessor-version":[{"id":38620,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38618\/revisions\/38620"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38618"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38618"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38618"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}