In one experiment, 2.50 grams of bornite is heated with excess oxygen, and 1.20 grams of copper is produced. What is the theoretical yield and percent yield for the experiment?<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n To determine the theoretical yield and percent yield of copper in this experiment, we first need to understand the chemical reaction involved in the heating of bornite (Cu\u2085FeS\u2084) with oxygen. The reaction is as follows: 2\u2009Cu5FeS4+11\u2009O2\u219210\u2009Cu+4\u2009Fe2O3+8\u2009SO22 \\, \\text{Cu}_5\\text{FeS}_4 + 11 \\, \\text{O}_2 \\rightarrow 10 \\, \\text{Cu} + 4 \\, \\text{Fe}_2\\text{O}_3 + 8 \\, \\text{SO}_2<\/p>\n\n\n\n First, we calculate the molar mass of bornite:<\/p>\n\n\n\n Bornite\u2019s formula is Cu\u2085FeS\u2084, so the molar mass is: (5\u00d763.55)+(1\u00d755.85)+(4\u00d732.07)=317.75+55.85+128.28=501.88\u2009g\/mol(5 \\times 63.55) + (1 \\times 55.85) + (4 \\times 32.07) = 317.75 + 55.85 + 128.28 = 501.88 \\, \\text{g\/mol}<\/p>\n\n\n\n Next, calculate the number of moles of bornite used: Moles of bornite=Mass of borniteMolar mass of bornite=2.50\u2009g501.88\u2009g\/mol=0.00498\u2009mol\\text{Moles of bornite} = \\frac{\\text{Mass of bornite}}{\\text{Molar mass of bornite}} = \\frac{2.50 \\, \\text{g}}{501.88 \\, \\text{g\/mol}} = 0.00498 \\, \\text{mol}<\/p>\n\n\n\n From the balanced equation, 2 moles of bornite produce 10 moles of copper. Therefore, the moles of copper produced from 0.00498 mol of bornite is: Moles of Cu=0.00498\u2009mol bornite\u00d710\u2009mol Cu2\u2009mol bornite=0.0249\u2009mol Cu\\text{Moles of Cu} = 0.00498 \\, \\text{mol bornite} \\times \\frac{10 \\, \\text{mol Cu}}{2 \\, \\text{mol bornite}} = 0.0249 \\, \\text{mol Cu}<\/p>\n\n\n\n Now, convert moles of copper to grams: Mass of Cu=Moles of Cu\u00d7Molar mass of Cu=0.0249\u2009mol\u00d763.55\u2009g\/mol=1.58\u2009g\\text{Mass of Cu} = \\text{Moles of Cu} \\times \\text{Molar mass of Cu} = 0.0249 \\, \\text{mol} \\times 63.55 \\, \\text{g\/mol} = 1.58 \\, \\text{g}<\/p>\n\n\n\n So, the theoretical yield of copper is 1.58 grams<\/strong>.<\/p>\n\n\n\n The percent yield is calculated using the formula: Percent yield=(Actual yieldTheoretical yield)\u00d7100=(1.20\u2009g1.58\u2009g)\u00d7100=75.95%\\text{Percent yield} = \\left( \\frac{\\text{Actual yield}}{\\text{Theoretical yield}} \\right) \\times 100 = \\left( \\frac{1.20 \\, \\text{g}}{1.58 \\, \\text{g}} \\right) \\times 100 = 75.95\\%<\/p>\n\n\n\n This experiment’s yield is relatively high, which suggests the process was efficient, but some loss or side reactions may have occurred, leading to a yield lower than the theoretical maximum.<\/p>\n","protected":false},"excerpt":{"rendered":" In one experiment, 2.50 grams of bornite is heated with excess oxygen, and 1.20 grams of copper is produced. What is the theoretical yield and percent yield for the experiment? The correct answer and explanation is: To determine the theoretical yield and percent yield of copper in this experiment, we first need to understand the […]<\/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-273725","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/273725","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=273725"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/273725\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=273725"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=273725"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=273725"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Molar Mass of Bornite (Cu\u2085FeS\u2084)<\/h3>\n\n\n\n
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Step 2: Moles of Bornite Used<\/h3>\n\n\n\n
Step 3: Theoretical Yield of Copper<\/h3>\n\n\n\n
Step 4: Percent Yield<\/h3>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
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