2H2 +O2 \u2192 2H2O
What mass of water forms when 1.45 \u00d7 10-3 g O2 react completely? (Molar mass of O2 = 32.00 g\/mol; molar mass of H2O = 18.02 g\/mol)<\/p>\n\n\n\n
1.63 \u00d7 10-3 g<\/p>\n\n\n\n
8.16 \u00d7 10-4 g<\/p>\n\n\n\n
1.29 \u00d7 10-3 g<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the mass of water ((H_2O)) that forms when 1.45 \u00d7 10^-3 g of (O_2) reacts completely, we start by using the balanced chemical equation:<\/p>\n\n\n\n [ From this equation, we see that one mole of (O_2) produces two moles of (H_2O).<\/p>\n\n\n\n To convert grams of (O_2) to moles, we use the formula:<\/p>\n\n\n\n [ The molar mass of (O_2) is 32.00 g\/mol. Thus,<\/p>\n\n\n\n [ According to the balanced equation, 1 mole of (O_2) produces 2 moles of (H_2O). Therefore, the moles of (H_2O) produced from 4.53 \u00d7 10^-5 moles of (O_2) is:<\/p>\n\n\n\n [ Now, we convert moles of (H_2O) to grams using its molar mass (18.02 g\/mol):<\/p>\n\n\n\n [ The mass of water that forms when 1.45 \u00d7 10^-3 g of (O_2) reacts completely is 1.63 \u00d7 10^-3 g<\/strong>. This value is obtained through stoichiometric calculations, ensuring that we accurately account for the mole ratios involved in the reaction.<\/p>\n","protected":false},"excerpt":{"rendered":" 2H2 +O2 \u2192 2H2OWhat mass of water forms when 1.45 \u00d7 10-3 g O2 react completely? (Molar mass of O2 = 32.00 g\/mol; molar mass of H2O = 18.02 g\/mol) 1.63 \u00d7 10-3 g 8.16 \u00d7 10-4 g 1.29 \u00d7 10-3 g The Correct Answer and Explanation is : To determine the mass of water […]<\/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-153902","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153902","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=153902"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153902\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=153902"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=153902"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=153902"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
2H_2 + O_2 \\rightarrow 2H_2O
]<\/p>\n\n\n\nStep 1: Convert the mass of (O_2) to moles.<\/h3>\n\n\n\n
\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}}
]<\/p>\n\n\n\n
\\text{moles of } O_2 = \\frac{1.45 \\times 10^{-3} \\text{ g}}{32.00 \\text{ g\/mol}} \\approx 4.53 \\times 10^{-5} \\text{ mol}
]<\/p>\n\n\n\nStep 2: Use the stoichiometry of the reaction to find moles of (H_2O).<\/h3>\n\n\n\n
\\text{moles of } H_2O = 2 \\times (4.53 \\times 10^{-5}) \\approx 9.06 \\times 10^{-5} \\text{ mol}
]<\/p>\n\n\n\nStep 3: Convert moles of (H_2O) to grams.<\/h3>\n\n\n\n
\\text{mass of } H_2O = \\text{moles} \\times \\text{molar mass} = 9.06 \\times 10^{-5} \\text{ mol} \\times 18.02 \\text{ g\/mol} \\approx 1.63 \\times 10^{-3} \\text{ g}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n