The molar mass of 2-chloro-2-methylbutane is 106.58 g\/mole, and its density is 0.866g\/ml. If 0.80 ml of 2-chloro-2-methylbutane were used in this experiment.
(a) In this experiment, moles of 2-chloro-2-methylbutane were used.
(b) The complete reaction of 0.80 ml. of 2-chloro-2-methylbutane would produce moles of HCI?
(c) To neutralize the total amount of HCl produced by the hydrolysis reaction, mL of 0.100M NaOH solution would be required.
(d) The 80% completion time of the reaction would require 0.100M NaOH solution. mL of the<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve the problem, we’ll go through each part of the question step-by-step:<\/p>\n\n\n\n First, we calculate the mass of 2-chloro-2-methylbutane: Now, convert the mass to moles: The reaction of 2-chloro-2-methylbutane in hydrolysis typically produces one mole of HCl per mole of 2-chloro-2-methylbutane. Thus, the moles of HCl produced would be equal to the moles of 2-chloro-2-methylbutane: To neutralize the HCl produced, we need an equivalent amount of NaOH since the reaction is ( \\text{NaOH} + \\text{HCl} \\to \\text{NaCl} + \\text{H}_2\\text{O} ). Thus: For 80% completion, we calculate 80% of the total moles of HCl expected and then calculate the volume of NaOH needed for that: These calculations help us understand the stoichiometry and reaction progression in typical organic chemistry hydrolysis reactions, emphasizing the importance of precise measurements and conversions in chemical experiments.<\/p>\n","protected":false},"excerpt":{"rendered":" The molar mass of 2-chloro-2-methylbutane is 106.58 g\/mole, and its density is 0.866g\/ml. If 0.80 ml of 2-chloro-2-methylbutane were used in this experiment.(a) In this experiment, moles of 2-chloro-2-methylbutane were used.(b) The complete reaction of 0.80 ml. of 2-chloro-2-methylbutane would produce moles of HCI?(c) To neutralize the total amount of HCl produced by the hydrolysis […]<\/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-194686","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194686","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=194686"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194686\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194686"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194686"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part (a): Calculating Moles of 2-chloro-2-methylbutane Used<\/h3>\n\n\n\n
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[ \\text{Mass} = \\text{Volume} \\times \\text{Density} = 0.80 \\, \\text{mL} \\times 0.866 \\, \\text{g\/mL} = 0.6928 \\, \\text{g} ]<\/p>\n\n\n\n
[ \\text{Moles} = \\frac{\\text{Mass}}{\\text{Molar mass}} = \\frac{0.6928 \\, \\text{g}}{106.58 \\, \\text{g\/mol}} = 0.00650 \\, \\text{mol} ]<\/p>\n\n\n\nPart (b): Moles of HCl Produced<\/h3>\n\n\n\n
[ \\text{Moles of HCl} = 0.00650 \\, \\text{mol} ]<\/p>\n\n\n\nPart (c): Volume of 0.100M NaOH Needed to Neutralize HCl<\/h3>\n\n\n\n
[ \\text{Volume of NaOH} = \\frac{\\text{Moles of HCl}}{\\text{Concentration of NaOH}} = \\frac{0.00650 \\, \\text{mol}}{0.100 \\, \\text{M\/L}} = 0.065 \\, \\text{L} = 65 \\, \\text{mL} ]<\/p>\n\n\n\nPart (d): Volume of 0.100M NaOH for 80% Completion<\/h3>\n\n\n\n
[ \\text{80% Moles of HCl} = 0.80 \\times 0.00650 \\, \\text{mol} = 0.00520 \\, \\text{mol} ]
[ \\text{Volume of NaOH for 80%} = \\frac{0.00520 \\, \\text{mol}}{0.100 \\, \\text{M\/L}} = 0.052 \\, \\text{L} = 52 \\, \\text{mL} ]<\/p>\n\n\n\nSummary<\/h3>\n\n\n\n
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