For the reaction represented by the equation SO3+H2O–>H22SO4 how many grams of sulfuric acid (H22SO4) can be produced from 197 g of water and 460g of sulfer trioxide?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine how many grams of sulfuric acid (H\u2082SO\u2084) can be produced from the given amounts of water (H\u2082O) and sulfur trioxide (SO\u2083), we need to follow the steps of stoichiometry. Let’s break it down step-by-step.<\/p>\n\n\n\n The balanced equation for the reaction is:<\/p>\n\n\n\n [ This indicates that one mole of sulfur trioxide (SO\u2083) reacts with one mole of water (H\u2082O) to produce one mole of sulfuric acid (H\u2082SO\u2084).<\/p>\n\n\n\n We are given 197 g of water (H\u2082O) and 460 g of sulfur trioxide (SO\u2083). Let’s calculate how many moles of each reactant we have:<\/p>\n\n\n\n From the balanced equation, we see that the stoichiometric ratio between SO\u2083 and H\u2082O is 1:1, so we need an equal number of moles of both reactants. Since we have fewer moles of SO\u2083 (5.75 mol) than H\u2082O (10.94 mol), SO\u2083<\/strong> is the limiting reactant.<\/p>\n\n\n\n Since SO\u2083 is the limiting reactant and the stoichiometric ratio is 1:1 (1 mole of SO\u2083 produces 1 mole of H\u2082SO\u2084), we will produce 5.75 moles of H\u2082SO\u2084.<\/p>\n\n\n\n Now, to find the mass of sulfuric acid produced, we multiply the moles of H\u2082SO\u2084 by its molar mass:<\/p>\n\n\n\n [ [ The maximum amount of sulfuric acid (H\u2082SO\u2084) that can be produced from 197 g of water and 460 g of sulfur trioxide is 563.5 grams<\/strong>.<\/p>\n\n\n\n For the reaction represented by the equation SO3+H2O–>H22SO4 how many grams of sulfuric acid (H22SO4) can be produced from 197 g of water and 460g of sulfer trioxide? The correct answer and explanation is : To determine how many grams of sulfuric acid (H\u2082SO\u2084) can be produced from the given amounts of water (H\u2082O) and […]<\/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-197791","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197791","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=197791"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197791\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197791"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197791"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197791"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Write the balanced chemical equation:<\/strong><\/h3>\n\n\n\n
SO_3 + H_2O \\rightarrow H_2SO_4
]<\/p>\n\n\n\n2. Calculate the molar masses of the substances involved:<\/strong><\/h3>\n\n\n\n
\n
H\u2082O consists of 2 hydrogen atoms and 1 oxygen atom.
Molar mass of H\u2082O = (2 \u00d7 1) + (1 \u00d7 16) = 18 g\/mol.<\/li>\n\n\n\n
SO\u2083 consists of 1 sulfur atom and 3 oxygen atoms.
Molar mass of SO\u2083 = (1 \u00d7 32) + (3 \u00d7 16) = 80 g\/mol.<\/li>\n\n\n\n
H\u2082SO\u2084 consists of 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms.
Molar mass of H\u2082SO\u2084 = (2 \u00d7 1) + (1 \u00d7 32) + (4 \u00d7 16) = 98 g\/mol.<\/li>\n<\/ul>\n\n\n\n3. Determine the limiting reactant:<\/strong><\/h3>\n\n\n\n
\n
[
\\text{Moles of H\u2082O} = \\frac{\\text{Mass of H\u2082O}}{\\text{Molar mass of H\u2082O}} = \\frac{197 \\, \\text{g}}{18 \\, \\text{g\/mol}} = 10.94 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Moles of SO\u2083} = \\frac{\\text{Mass of SO\u2083}}{\\text{Molar mass of SO\u2083}} = \\frac{460 \\, \\text{g}}{80 \\, \\text{g\/mol}} = 5.75 \\, \\text{mol}
]<\/li>\n<\/ul>\n\n\n\n4. Calculate the moles of H\u2082SO\u2084 produced:<\/strong><\/h3>\n\n\n\n
5. Convert moles of H\u2082SO\u2084 to grams:<\/strong><\/h3>\n\n\n\n
\\text{Mass of H\u2082SO\u2084} = \\text{Moles of H\u2082SO\u2084} \\times \\text{Molar mass of H\u2082SO\u2084}
]<\/p>\n\n\n\n
\\text{Mass of H\u2082SO\u2084} = 5.75 \\, \\text{mol} \\times 98 \\, \\text{g\/mol} = 563.5 \\, \\text{g}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
\n
This gives us the final mass of sulfuric acid, which is 563.5 grams.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"