How many moles of aluminum chloride ALCL3 for when 81 g AL reacts with plenty of chlorine? (AI: 27 g\/mol) 2AI + 3CI2 —> 2AICI3<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine how many moles of aluminum chloride (AlCl\u2083) are produced when 81 grams of aluminum (Al) reacts with plenty of chlorine gas (Cl\u2082), we first need to start with the balanced chemical equation:<\/p>\n\n\n\n [ Step 1: Calculate the number of moles of aluminum (Al)<\/strong><\/p>\n\n\n\n The molar mass of aluminum (Al) is given as 27 g\/mol. To find the number of moles of aluminum, we use the formula:<\/p>\n\n\n\n [ Substituting in the given values:<\/p>\n\n\n\n [ Step 2: Use the stoichiometry of the reaction<\/strong><\/p>\n\n\n\n From the balanced equation, we see that 2 moles of Al react to produce 2 moles of AlCl\u2083. This means that the molar ratio of Al to AlCl\u2083 is 1:1. Therefore, for every mole of aluminum, one mole of aluminum chloride is produced.<\/p>\n\n\n\n Since we have calculated that there are 3 moles of aluminum, we can conclude that:<\/p>\n\n\n\n [ Thus, when 81 grams of aluminum reacts with plenty of chlorine gas, it produces 3 moles of aluminum chloride (AlCl\u2083)<\/strong>.<\/p>\n\n\n\n Conclusion:<\/strong><\/p>\n\n\n\n In summary, starting with 81 grams of aluminum, we first calculated the moles of aluminum using its molar mass, finding that it yields 3 moles of Al. Based on the stoichiometry of the reaction, this results in the formation of 3 moles of AlCl\u2083. The balanced equation allows us to directly correlate the amount of reactant to the product, ensuring that we understand the precise quantities involved in the chemical reaction.<\/p>\n","protected":false},"excerpt":{"rendered":" How many moles of aluminum chloride ALCL3 for when 81 g AL reacts with plenty of chlorine? (AI: 27 g\/mol) 2AI + 3CI2 —> 2AICI3 The Correct Answer and Explanation is : To determine how many moles of aluminum chloride (AlCl\u2083) are produced when 81 grams of aluminum (Al) reacts with plenty of chlorine gas […]<\/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-153440","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153440","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=153440"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153440\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=153440"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=153440"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=153440"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
2 \\text{Al} + 3 \\text{Cl}_2 \\rightarrow 2 \\text{AlCl}_3
]<\/p>\n\n\n\n
\\text{moles of Al} = \\frac{\\text{mass of Al}}{\\text{molar mass of Al}}
]<\/p>\n\n\n\n
\\text{moles of Al} = \\frac{81 \\, \\text{g}}{27 \\, \\text{g\/mol}} = 3 \\, \\text{moles of Al}
]<\/p>\n\n\n\n
\\text{moles of AlCl}_3 = 3 \\, \\text{moles of Al}
]<\/p>\n\n\n\n