{"id":32519,"date":"2025-06-22T15:05:15","date_gmt":"2025-06-22T15:05:15","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32519"},"modified":"2025-06-22T15:05:16","modified_gmt":"2025-06-22T15:05:16","slug":"determine-the-percent-composition-of-each-element-in-nh4br-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-percent-composition-of-each-element-in-nh4br-2\/","title":{"rendered":"Determine the percent composition of each element in NH4Br"},"content":{"rendered":"\n<p>Determine the percent composition of each element in NH4Br. It should be rounded to the nearest percent. 2.) 3.25 x 10^25 atoms of neon gas (Ne) equals how many moles of neon gas (Ne)? 3.) How many moles of Al can be produced from 10.87g of Ag? Al(NO3)3(s) + 3Ag &#8212;> Al + 3AgNO3 4.) How many grams of N2 are needed to produce 2.17 mol of NH3 when reacted according to this chemical equation? N2(g) + 3H2(g) &#8212;> 2NH3(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<h3 class=\"wp-block-heading\"><strong>1. Percent Composition of NH\u2084Br<\/strong><\/h3>\n\n\n\n<p><strong>Molar masses:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>N = 14 g\/mol<\/li>\n\n\n\n<li>H = 1 g\/mol \u00d7 4 = 4 g\/mol<\/li>\n\n\n\n<li>Br = 79.9 g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Molar mass of NH\u2084Br = 14 + 4 + 79.9 = 97.9 g\/mol<\/strong><\/p>\n\n\n\n<p><strong>Percent Composition:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>N: (14 \/ 97.9) \u00d7 100 \u2248 14%<\/li>\n\n\n\n<li>H: (4 \/ 97.9) \u00d7 100 \u2248 4%<\/li>\n\n\n\n<li>Br: (79.9 \/ 97.9) \u00d7 100 \u2248 82%<\/li>\n<\/ul>\n\n\n\n<p><strong>Answer:<\/strong><br>Nitrogen = 14%, Hydrogen = 4%, Bromine = 82%<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Moles from Atoms of Neon<\/strong><\/h3>\n\n\n\n<p>Use Avogadro\u2019s number:<br>1 mole = 6.022 \u00d7 10\u00b2\u00b3 atoms<\/p>\n\n\n\n<p><strong>Given:<\/strong> 3.25 \u00d7 10\u00b2\u2075 atoms Ne<br><strong>Calculation:<\/strong><br>Moles = (3.25 \u00d7 10\u00b2\u2075 atoms) \/ (6.022 \u00d7 10\u00b2\u00b3 atoms\/mol) \u2248 54.0 mol<\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>54 moles of Ne<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>3. Moles of Al from 10.87 g of Ag<\/strong><\/h3>\n\n\n\n<p><strong>Balanced equation:<\/strong><br>Al(NO\u2083)\u2083 + 3Ag \u2192 Al + 3AgNO\u2083<\/p>\n\n\n\n<p><strong>Molar mass of Ag = 107.87 g\/mol<\/strong><br><strong>Given mass = 10.87 g<\/strong><\/p>\n\n\n\n<p><strong>Moles of Ag = 10.87 \/ 107.87 \u2248 0.101 mol<\/strong><\/p>\n\n\n\n<p><strong>From stoichiometry:<\/strong><br>3 mol Ag \u2192 1 mol Al<br>So,<br>0.101 mol Ag \u00d7 (1 mol Al \/ 3 mol Ag) \u2248 0.0337 mol Al<\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>0.034 moles of Al<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>4. Grams of N\u2082 to Produce 2.17 mol NH\u2083<\/strong><\/h3>\n\n\n\n<p><strong>Balanced equation:<\/strong><br>N\u2082 + 3H\u2082 \u2192 2NH\u2083<\/p>\n\n\n\n<p><strong>From stoichiometry:<\/strong><br>2 mol NH\u2083 \u2192 1 mol N\u2082<br>So,<br>2.17 mol NH\u2083 \u00d7 (1 mol N\u2082 \/ 2 mol NH\u2083) = 1.085 mol N\u2082<\/p>\n\n\n\n<p><strong>Molar mass of N\u2082 = 28.02 g\/mol<\/strong><\/p>\n\n\n\n<p><strong>Mass = 1.085 mol \u00d7 28.02 g\/mol \u2248 30.4 g<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>30 grams of N\u2082<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (Summary of Concepts Used)<\/strong><\/h3>\n\n\n\n<p>These problems use key chemistry concepts: percent composition, mole conversions, and stoichiometry. Percent composition requires dividing each element\u2019s mass by the total molar mass and multiplying by 100. Converting atoms to moles uses Avogadro\u2019s number. Stoichiometry problems follow balanced chemical equations and molar ratios. For mass-to-mole or mole-to-mass conversions, the molar mass from the periodic table bridges between grams and moles. Each solution here ensures correct scientific methodology and proper rounding to the nearest whole number where needed.<\/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-banner8-396.jpeg\" alt=\"\" class=\"wp-image-32520\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-396.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-396-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-396-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the percent composition of each element in NH4Br. It should be rounded to the nearest percent. 2.) 3.25 x 10^25 atoms of neon gas (Ne) equals how many moles of neon gas (Ne)? 3.) How many moles of Al can be produced from 10.87g of Ag? Al(NO3)3(s) + 3Ag &#8212;> Al + 3AgNO3 4.) [&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-32519","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32519","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=32519"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32519\/revisions"}],"predecessor-version":[{"id":32521,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32519\/revisions\/32521"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32519"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32519"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32519"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}