5 Ml Of Bromo-2-Methylpropane Were Used, Density Of Brono-2-Methylpropane Is 1.22 G\/ML<\/p>\n\n\n\n
Calculate the mass of naphthalene required to react stoichiometrically with the moles of 2- bromo-2-methylpropane that you have placed in the flask. Dispense that quantity of naphthalene into the weighing dish. There is some imprecision in delivering the powder onto the weigh boat, but a mass within 0.040 g of the calculated stoichiometric quantity will be sufficient for this experiment. Note that you can use a combination of the 1 g and the 0.1 g buttons to add the required mass. If you overshoot the goal, you can discard the weighing boat and get a new one. 5 ml of bromo-2-methylpropane RECord the sample mass dispensed below mass of naphthalene (g) 5.9924 Move the used scoopula into the Recycle cabinet.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the mass of naphthalene required to react stoichiometrically with 5 mL of bromo-2-methylpropane, we need to go through the following steps:<\/p>\n\n\n\n We are given the volume and density of bromo-2-methylpropane:<\/p>\n\n\n\n The mass of bromo-2-methylpropane can be calculated using the formula: To find the moles of bromo-2-methylpropane, we need the molar mass of bromo-2-methylpropane. Assuming the chemical formula for bromo-2-methylpropane is ( \\text{C}_4\\text{H}_9\\text{Br} ):<\/p>\n\n\n\n So, the molar mass of bromo-2-methylpropane is: Now, calculate the moles of bromo-2-methylpropane: From the balanced chemical equation for the reaction between bromo-2-methylpropane and naphthalene, assuming a 1:1 mole ratio (since no specific stoichiometry is provided): Now, we need the molar mass of naphthalene (( \\text{C}_10\\text{H}_8 )):<\/p>\n\n\n\n Thus, the molar mass of naphthalene is: Now, calculate the mass of naphthalene: The mass of naphthalene required to react with 5 mL of bromo-2-methylpropane is approximately 5.70 g<\/strong>. Since there is some imprecision in delivering the powder, a mass within 0.040 g of this value is acceptable. Therefore, the mass of naphthalene dispensed (5.9924 g) is slightly above the stoichiometric requirement, but this is within an acceptable range.<\/p>\n","protected":false},"excerpt":{"rendered":" 5 Ml Of Bromo-2-Methylpropane Were Used, Density Of Brono-2-Methylpropane Is 1.22 G\/ML Calculate the mass of naphthalene required to react stoichiometrically with the moles of 2- bromo-2-methylpropane that you have placed in the flask. Dispense that quantity of naphthalene into the weighing dish. There is some imprecision in delivering the powder onto the weigh boat, […]<\/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-181999","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181999","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=181999"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181999\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181999"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181999"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181999"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate the mass of bromo-2-methylpropane<\/h3>\n\n\n\n
\n
[
\\text{Mass} = \\text{Density} \\times \\text{Volume}
]
[
\\text{Mass of bromo-2-methylpropane} = 1.22 \\, \\text{g\/mL} \\times 5 \\, \\text{mL} = 6.10 \\, \\text{g}
]<\/p>\n\n\n\nStep 2: Calculate the moles of bromo-2-methylpropane<\/h3>\n\n\n\n
\n
[
\\text{Molar mass} = 48.04 + 9.072 + 79.904 = 137.016 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
[
\\text{Moles of bromo-2-methylpropane} = \\frac{\\text{Mass}}{\\text{Molar mass}} = \\frac{6.10 \\, \\text{g}}{137.016 \\, \\text{g\/mol}} = 0.0445 \\, \\text{mol}
]<\/p>\n\n\n\nStep 3: Calculate the moles of naphthalene required<\/h3>\n\n\n\n
[
\\text{Moles of naphthalene} = 0.0445 \\, \\text{mol}
]<\/p>\n\n\n\nStep 4: Calculate the mass of naphthalene required<\/h3>\n\n\n\n
\n
[
\\text{Molar mass of naphthalene} = 120.10 + 8.064 = 128.164 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
[
\\text{Mass of naphthalene} = \\text{Moles of naphthalene} \\times \\text{Molar mass of naphthalene} = 0.0445 \\, \\text{mol} \\times 128.164 \\, \\text{g\/mol} = 5.70 \\, \\text{g}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n