How many grams of oxygen are required to produce 100 mols of water H2O in a synthesis reaction with excess hydrogen?<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n To find out how many grams of oxygen are required to produce 100 moles of water (H\u2082O), we need to understand the stoichiometry of the synthesis reaction between hydrogen and oxygen:<\/p>\n\n\n\n The balanced chemical equation for the formation of water is: From the equation, we can see that 1 mole of O\u2082 produces 2 moles of H\u2082O.<\/p>\n\n\n\n If 2 moles of H\u2082O are produced from 1 mole of O\u2082, then to produce 100 moles of H\u2082O, we can use a simple proportion: 1 mole of O\u20822 moles of H\u2082O=x moles of O\u2082100 moles of H\u2082O\\frac{1 \\text{ mole of O\u2082}}{2 \\text{ moles of H\u2082O}} = \\frac{x \\text{ moles of O\u2082}}{100 \\text{ moles of H\u2082O}}<\/p>\n\n\n\n Solving for x: x=100\u00d712=50 moles of O\u2082x = \\frac{100 \\times 1}{2} = 50 \\text{ moles of O\u2082}<\/p>\n\n\n\n Thus, 50 moles of oxygen are required to produce 100 moles of water.<\/p>\n\n\n\n The molar mass of oxygen (O\u2082) is approximately 32 grams per mole. To find the mass of 50 moles of oxygen, we multiply the number of moles by the molar mass: Mass of O\u2082=50 moles\u00d732 g\/mol=1600 grams of O\u2082\\text{Mass of O\u2082} = 50 \\text{ moles} \\times 32 \\text{ g\/mol} = 1600 \\text{ grams of O\u2082}<\/p>\n\n\n\n To produce 100 moles of water, 1600 grams of oxygen are required.<\/p>\n\n\n\n This calculation shows how stoichiometry helps in converting moles of a product into the required amount of reactant. The key is balancing the chemical equation and using the molar relationships between reactants and products.<\/p>\n","protected":false},"excerpt":{"rendered":" How many grams of oxygen are required to produce 100 mols of water H2O in a synthesis reaction with excess hydrogen? The correct answer and explanation is: To find out how many grams of oxygen are required to produce 100 moles of water (H\u2082O), we need to understand the stoichiometry of the synthesis reaction between […]<\/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-245184","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245184","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=245184"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245184\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245184"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245184"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245184"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
2H2(g)+O2(g)\u21922H2O(l)2H_2 (g) + O_2 (g) \\rightarrow 2H_2O (l)<\/p>\n\n\n\nStep 1: Moles of Oxygen Needed<\/h3>\n\n\n\n
Step 2: Mass of Oxygen Required<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n