To solve this problem, we need to determine the limiting reagent in the reaction and then calculate how much sulfuric acid (H\u2082SO\u2084) can be produced from the given amounts of reactants. The balanced chemical equation is:<\/p>\n\n\n\n
SO3+H2O\u2192H2SO4SO_3 + H_2O \\rightarrow H_2SO_4SO3\u200b+H2\u200bO\u2192H2\u200bSO4\u200b<\/p>\n\n\n\n
Step 1: Molar Masses<\/h3>\n\n\n\n
First, we need to know the molar masses of the compounds involved:<\/p>\n\n\n\n
- \n
- Molar mass of H\u2082O (water)<\/strong> = 18.02 g\/mol<\/li>\n\n\n\n
- Molar mass of SO\u2083 (sulfur trioxide)<\/strong> = 80.06 g\/mol<\/li>\n\n\n\n
- Molar mass of H\u2082SO\u2084 (sulfuric acid)<\/strong> = 98.08 g\/mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Moles of Reactants<\/h3>\n\n\n\n
Next, we calculate how many moles of each reactant are available.<\/p>\n\n\n\n
Moles of H\u2082O:<\/h4>\n\n\n\n
Given that we have 320 g of water:moles of H2O=320\u2009g18.02\u2009g\/mol\u224817.75\u2009mol\\text{moles of H}_2O = \\frac{320 \\, \\text{g}}{18.02 \\, \\text{g\/mol}} \\approx 17.75 \\, \\text{mol}moles of H2\u200bO=18.02g\/mol320g\u200b\u224817.75mol<\/p>\n\n\n\n
Moles of SO\u2083:<\/h4>\n\n\n\n
Given that we have 1064 g of sulfur trioxide:moles of SO3=1064\u2009g80.06\u2009g\/mol\u224813.30\u2009mol\\text{moles of SO}_3 = \\frac{1064 \\, \\text{g}}{80.06 \\, \\text{g\/mol}} \\approx 13.30 \\, \\text{mol}moles of SO3\u200b=80.06g\/mol1064g\u200b\u224813.30mol<\/p>\n\n\n\n
Step 3: Identify the Limiting Reagent<\/h3>\n\n\n\n
From the balanced equation, we see that the mole ratio of SO\u2083 to H\u2082O is 1:1. This means that for every 1 mole of SO\u2083, 1 mole of H\u2082O is required to produce H\u2082SO\u2084.<\/p>\n\n\n\n
- \n
- We have 17.75 moles of H\u2082O and 13.30 moles of SO\u2083.<\/li>\n\n\n\n
- Since SO\u2083 is the limiting reagent (because we have fewer moles of SO\u2083), the reaction will be limited by the amount of SO\u2083.<\/li>\n<\/ul>\n\n\n\n
Step 4: Calculate the Grams of H\u2082SO\u2084 Produced<\/h3>\n\n\n\n
According to the reaction, 1 mole of SO\u2083 produces 1 mole of H\u2082SO\u2084. Therefore, the moles of H\u2082SO\u2084 produced will be equal to the moles of SO\u2083.moles of H2SO4=13.30\u2009mol\\text{moles of H}_2SO_4 = 13.30 \\, \\text{mol}moles of H2\u200bSO4\u200b=13.30mol<\/p>\n\n\n\n
Now, calculate the mass of sulfuric acid produced:mass of H2SO4=moles of H2SO4\u00d7molar mass of H2SO4\\text{mass of H}_2SO_4 = \\text{moles of H}_2SO_4 \\times \\text{molar mass of H}_2SO_4mass of H2\u200bSO4\u200b=moles of H2\u200bSO4\u200b\u00d7molar mass of H2\u200bSO4\u200bmass of H2SO4=13.30\u2009mol\u00d798.08\u2009g\/mol\u22481303.26\u2009g\\text{mass of H}_2SO_4 = 13.30 \\, \\text{mol} \\times 98.08 \\, \\text{g\/mol} \\approx 1303.26 \\, \\text{g}mass of H2\u200bSO4\u200b=13.30mol\u00d798.08g\/mol\u22481303.26g<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The mass of sulfuric acid that can be produced is approximately 1303.26 g<\/strong>.<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"For the reaction represented by the equation SO3 + H2O -> H2SO4, how many grams of sulfuric acid can be produced from 320 g of water and 1064 g of sulfuric trioxide? The Correct Answer and Explanation is: To solve this problem, we need to determine the limiting reagent in the reaction and then calculate […]<\/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-258151","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258151","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=258151"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258151\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258151"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258151"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Molar mass of SO\u2083 (sulfur trioxide)<\/strong> = 80.06 g\/mol<\/li>\n\n\n\n