The neutralization reaction between hydrochloric acid and calcium hydroxide produces calcium chloride and water. 2 HCl + Ca(OH)2 = 2 H2O + CaCl2 If 0.75 grams of Ca(OH)2 is used in the reaction, determine the following: a. Number of moles of HCl(MM=36.46 g\/mol) needed to completely react with Ca(OH)2. The molar mass of Ca(OH)2 is 74.10 g\/mol. b. Mass of CaCl2(MM=110.98 g\/mol) formed in the reaction.
The neutralization reaction between hydrochloric acid and calcium hydroxide produces calcium chloride and water.
2 HCl + Ca(OH)2 = 2 H2O + CaCl2<\/p>\n\n\n\n
If 0.75 grams of Ca(OH)2 is used in the reaction, determine the following:<\/p>\n\n\n\n
a. Number of moles of HCl(MM=36.46 g\/mol) needed to completely react with Ca(OH)2. The molar mass of Ca(OH)2 is 74.10 g\/mol.<\/p>\n\n\n\n
b. Mass of CaCl2(MM=110.98 g\/mol) formed in the reaction.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve the problem, let’s address each part step-by-step.<\/p>\n\n\n\n a. Moles of ( \\text{HCl} ) needed:<\/strong> ( 0.02024 \\, \\text{mol} )<\/p>\n\n\n\n b. Mass of ( \\text{CaCl}_2 ) formed:<\/strong> ( 1.123 \\, \\text{g} )<\/p>\n\n\n\n The neutralization reaction is a double displacement reaction between hydrochloric acid and calcium hydroxide. Based on the molar ratio from the balanced chemical equation, ( \\text{Ca(OH)}_2 ) reacts with twice the moles of ( \\text{HCl} ), and the amount of ( \\text{CaCl}_2 ) produced is directly proportional to the moles of ( \\text{Ca(OH)}_2 ). Using the given masses and molar masses, we converted grams to moles, used stoichiometric relationships, and calculated the required quantities step-by-step. This method ensures accurate results consistent with the law of conservation of mass.<\/p>\n","protected":false},"excerpt":{"rendered":" The neutralization reaction between hydrochloric acid and calcium hydroxide produces calcium chloride and water. 2 HCl + Ca(OH)2 = 2 H2O + CaCl2 If 0.75 grams of Ca(OH)2 is used in the reaction, determine the following: a. Number of moles of HCl(MM=36.46 g\/mol) needed to completely react with Ca(OH)2. The molar mass of Ca(OH)2 is […]<\/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-184891","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184891","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=184891"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184891\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184891"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184891"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184891"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given Data:<\/h3>\n\n\n\n
\n
\n\n\n\na. Moles of ( \\text{HCl} ) Required:<\/h3>\n\n\n\n
\n
[
\\text{Moles of } \\text{Ca(OH)}_2 = \\frac{\\text{Mass of } \\text{Ca(OH)}_2}{\\text{Molar mass of } \\text{Ca(OH)}_2}
]
[
\\text{Moles of } \\text{Ca(OH)}_2 = \\frac{0.75 \\, \\text{g}}{74.10 \\, \\text{g\/mol}} = 0.01012 \\, \\text{mol}
]<\/li>\n\n\n\n
The balanced equation shows that 1 mole of ( \\text{Ca(OH)}_2 ) reacts with 2 moles of ( \\text{HCl} ). Thus:
[
\\text{Moles of } \\text{HCl} = 0.01012 \\, \\text{mol} \\times 2 = 0.02024 \\, \\text{mol}
]<\/li>\n<\/ol>\n\n\n\n
\n\n\n\nb. Mass of ( \\text{CaCl}_2 ) Formed:<\/h3>\n\n\n\n
\n
From the balanced equation, 1 mole of ( \\text{Ca(OH)}_2 ) produces 1 mole of ( \\text{CaCl}_2 ). Therefore:
[
\\text{Moles of } \\text{CaCl}_2 = 0.01012 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{Mass of } \\text{CaCl}_2 = \\text{Moles of } \\text{CaCl}_2 \\times \\text{Molar mass of } \\text{CaCl}_2
]
[
\\text{Mass of } \\text{CaCl}_2 = 0.01012 \\, \\text{mol} \\times 110.98 \\, \\text{g\/mol} = 1.123 \\, \\text{g}
]<\/li>\n<\/ol>\n\n\n\n
\n\n\n\nFinal Answers:<\/h3>\n\n\n\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n