{"id":238182,"date":"2025-06-17T21:13:44","date_gmt":"2025-06-17T21:13:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=238182"},"modified":"2025-06-17T21:13:46","modified_gmt":"2025-06-17T21:13:46","slug":"consider-the-following-unbalanced-equation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/17\/consider-the-following-unbalanced-equation\/","title":{"rendered":"consider the following unbalanced equation."},"content":{"rendered":"\n

consider the following unbalanced equation. ——– Mg + —– O2 \u2192 —– MgO Balance the chemical equation, and how many grams of MgO are produced when 43.0 grams of O2 reacts completely with excess Mg? some possibly useful molar masses are as follows: O2 = 32.00 g\/mol , MgO = 40.30 g\/mol<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Balanced Chemical Equation:<\/strong><\/h3>\n\n\n\n

2Mg+O2\u21922MgO2 \\text{Mg} + \\text{O}_2 \\rightarrow 2 \\text{MgO}2Mg+O2\u200b\u21922MgO<\/p>\n\n\n\n

Step-by-step Solution:<\/strong><\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • Mass of O2\\text{O}_2O2\u200b = 43.0 g<\/li>\n\n\n\n
  • Molar mass of O2\\text{O}_2O2\u200b = 32.00 g\/mol<\/li>\n\n\n\n
  • Molar mass of MgO\\text{MgO}MgO = 40.30 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Step 1: Convert grams of O\u2082 to moles<\/strong><\/h4>\n\n\n\n

    Moles of O2=43.0 g32.00 g\/mol=1.34375 mol\\text{Moles of } O_2 = \\frac{43.0\\ \\text{g}}{32.00\\ \\text{g\/mol}} = 1.34375\\ \\text{mol}Moles of O2\u200b=32.00 g\/mol43.0 g\u200b=1.34375 mol<\/p>\n\n\n\n

    Step 2: Use stoichiometry<\/strong><\/h4>\n\n\n\n

    From the balanced equation, 1 mole of O2\\text{O}_2O2\u200b produces 2 moles of MgO\\text{MgO}MgO.Moles of MgO=1.34375 mol O2\u00d72 mol MgO1 mol O2=2.6875 mol MgO\\text{Moles of } MgO = 1.34375\\ \\text{mol } O_2 \\times \\frac{2\\ \\text{mol } MgO}{1\\ \\text{mol } O_2} = 2.6875\\ \\text{mol } MgOMoles of MgO=1.34375 mol O2\u200b\u00d71 mol O2\u200b2 mol MgO\u200b=2.6875 mol MgO<\/p>\n\n\n\n

    Step 3: Convert moles of MgO to grams<\/strong><\/h4>\n\n\n\n

    Mass of MgO=2.6875 mol\u00d740.30 g\/mol=108.27 g\\text{Mass of } MgO = 2.6875\\ \\text{mol} \\times 40.30\\ \\text{g\/mol} = 108.27\\ \\text{g}Mass of MgO=2.6875 mol\u00d740.30 g\/mol=108.27 g<\/p>\n\n\n\n

    Final Answer:<\/strong><\/h3>\n\n\n\n

    108.27 grams of MgO<\/strong> are produced when 43.0 grams of O\u2082 reacts completely with excess Mg.<\/p>\n\n\n\n

    \n\n\n\n

    Explanation <\/strong><\/h3>\n\n\n\n

    Balancing chemical equations is essential for understanding how much of each substance participates in a reaction. The unbalanced equation given is:Mg+O2\u2192MgO\\text{Mg} + \\text{O}_2 \\rightarrow \\text{MgO}Mg+O2\u200b\u2192MgO<\/p>\n\n\n\n

    Oxygen exists as a diatomic molecule (O2\\text{O}_2O2\u200b), so to balance the equation, we must ensure equal numbers of atoms on both sides. Since one O2\\text{O}_2O2\u200b molecule provides two oxygen atoms, and each MgO compound contains one oxygen atom, we need two Mg atoms and two MgO molecules:2Mg+O2\u21922MgO2 \\text{Mg} + \\text{O}_2 \\rightarrow 2 \\text{MgO}2Mg+O2\u200b\u21922MgO<\/p>\n\n\n\n

    Next, we calculate how much MgO is produced from 43.0 grams of O2\\text{O}_2O2\u200b. First, we convert grams to moles using the molar mass of oxygen gas, which is 32.00 g\/mol. This gives us approximately 1.344 moles of O2\\text{O}_2O2\u200b. According to the balanced equation, every mole of O2\\text{O}_2O2\u200b produces 2 moles of MgO. Multiplying gives us 2.688 moles of MgO.<\/p>\n\n\n\n

    Finally, we convert moles of MgO to grams using its molar mass of 40.30 g\/mol. Multiplying the moles by the molar mass gives us approximately 108.27 grams.<\/p>\n\n\n\n

    This calculation assumes magnesium is in excess, meaning all the oxygen is used up in the reaction. Therefore, the limiting reactant is oxygen, and the final mass of magnesium oxide is based solely on the available oxygen.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    consider the following unbalanced equation. ——– Mg + —– O2 \u2192 —– MgO Balance the chemical equation, and how many grams of MgO are produced when 43.0 grams of O2 reacts completely with excess Mg? some possibly useful molar masses are as follows: O2 = 32.00 g\/mol , MgO = 40.30 g\/mol The Correct Answer […]<\/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-238182","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238182","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=238182"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238182\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238182"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238182"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238182"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}