Gas Stoichiometry Worksheet Name: Solve all the following gas law problems. Show all work, answers are given at the end of the problem. Molar Volume 1. Calculate the number of moles contained in 550.mL of carbon dioxide at STP. (0.0246mol) 2. Calculate the mass of 1.50 L of CH4 at STP. (1.07g) 3. Calculate the volume in liters of 50.0 grams of nitrogen dioxide at STP. (24.3L) Gas Stoichiometry 4. Given the following reaction: NH.NO, (s) + N2(g) + 2H20 (g) How many liters of nitrogen gas is produced if 50.0L of water is produced at STP? (25.0L) 5. Given the following reaction: CuO (s) + H2(g) ? Cu (s) + H20 (g) If 250. L of hydrogen gas are used to reduce copper (II) oxide at STP, what mass of copper is to be expected? (7095) 6. How many grams of solid lithium must be added to liquid water in order to obtain 15.0 L of hydrogen gas at STP? (9.29g) 2 Li + 21,0 + 2LiOH + H2<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go through each of the gas law problems step by step, showing the work and explaining the calculations.<\/p>\n\n\n\n We use the molar volume at STP (Standard Temperature and Pressure) where 1 mole of gas occupies 22.4 L (or 22.4 \u00d7 10\u00b3 mL).<\/p>\n\n\n\n Given:<\/strong><\/p>\n\n\n\n Formula:<\/strong> To calculate the mass, we need the molar mass of CH\u2084, which is 16.04 g\/mol. Using the molar volume at STP:<\/p>\n\n\n\n Given:<\/strong><\/p>\n\n\n\n Steps:<\/strong><\/p>\n\n\n\n To solve this, first, find the moles of NO\u2082, then use the molar volume to calculate the volume.<\/p>\n\n\n\n Given:<\/strong><\/p>\n\n\n\n Steps:<\/strong><\/p>\n\n\n\n The reaction is: Given:<\/strong><\/p>\n\n\n\n From the balanced equation:<\/strong><\/p>\n\n\n\n [ The reaction is: Given:<\/strong><\/p>\n\n\n\n Steps:<\/strong><\/p>\n\n\n\n The reaction is: Given:<\/strong><\/p>\n\n\n\n Steps:<\/strong><\/p>\n\n\n\n Gas Stoichiometry Worksheet Name: Solve all the following gas law problems. Show all work, answers are given at the end of the problem. Molar Volume 1. Calculate the number of moles contained in 550.mL of carbon dioxide at STP. (0.0246mol) 2. Calculate the mass of 1.50 L of CH4 at STP. (1.07g) 3. Calculate the […]<\/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-189207","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189207","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=189207"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189207\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189207"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189207"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189207"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Number of moles in 550 mL of CO\u2082 at STP<\/strong><\/h3>\n\n\n\n
\n
[
\\text{Moles of gas} = \\frac{\\text{Volume of gas}}{\\text{Molar Volume}}
]
[
\\text{Moles of CO\u2082} = \\frac{0.550}{22.4} = 0.0246 \\, \\text{mol}
]<\/p>\n\n\n\n2. Mass of 1.50 L of CH\u2084 at STP<\/strong><\/h3>\n\n\n\n
\n
\n
[
\\text{Moles of CH\u2084} = \\frac{1.50}{22.4} = 0.0670 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Mass} = 0.0670 \\times 16.04 = 1.07 \\, \\text{g}
]<\/li>\n<\/ol>\n\n\n\n3. Volume of 50.0 grams of NO\u2082 at STP<\/strong><\/h3>\n\n\n\n
\n
\n
[
\\text{Moles of NO\u2082} = \\frac{50.0}{46.0} = 1.087 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Volume} = 1.087 \\times 22.4 = 24.3 \\, \\text{L}
]<\/li>\n<\/ol>\n\n\n\n4. Liters of N\u2082 produced if 50.0 L of H\u2082O is produced<\/strong><\/h3>\n\n\n\n
[
\\text{NH\u2084NO\u2083 (s)} \\rightarrow \\text{N\u2082(g)} + 2 \\text{H\u2082O (g)}
]
From the balanced equation, 1 mole of NH\u2084NO\u2083 produces 1 mole of N\u2082 and 2 moles of H\u2082O. If 50.0 L of H\u2082O is produced, we can use the molar volume at STP to relate volumes of gases.<\/p>\n\n\n\n\n
\n
\\text{Volume of N\u2082} = \\frac{50.0}{2} = 25.0 \\, \\text{L}
]<\/p>\n\n\n\n5. Mass of Copper produced from 250 L of H\u2082 at STP<\/strong><\/h3>\n\n\n\n
[
\\text{CuO (s)} + \\text{H\u2082 (g)} \\rightarrow \\text{Cu (s)} + \\text{H\u2082O (g)}
]
From the balanced equation, 1 mole of H\u2082 reacts to form 1 mole of Cu. We can use the molar volume to find the moles of H\u2082 and then the mass of Cu.<\/p>\n\n\n\n\n
\n
[
\\text{Moles of H\u2082} = \\frac{250}{22.4} = 11.16 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Moles of Cu} = 11.16 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Mass of Cu} = 11.16 \\times 63.5 = 709.5 \\, \\text{g}
]<\/li>\n<\/ol>\n\n\n\n6. Grams of Lithium needed to obtain 15.0 L of H\u2082 at STP<\/strong><\/h3>\n\n\n\n
[
2 \\text{Li} + 2 \\text{H\u2082O} \\rightarrow 2 \\text{LiOH} + \\text{H\u2082}
]
From the balanced equation, 2 moles of Li produce 1 mole of H\u2082. We can use the molar volume to find moles of H\u2082 and then calculate the mass of lithium required.<\/p>\n\n\n\n\n
\n
[
\\text{Moles of H\u2082} = \\frac{15.0}{22.4} = 0.669 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Moles of Li} = 0.669 \\times 2 = 1.338 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Mass of Li} = 1.338 \\times 6.94 = 9.29 \\, \\text{g}
]<\/li>\n<\/ol>\n\n\n\nSummary of answers:<\/h3>\n\n\n\n
\n