Calcium carbonate reacts with aqueous HCl to give CaCl2and CO2according to the reaction given below: CaCO3(s) + 2HCl (aq) ? CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2will be formed when 250 mL of 0.76 M HCI reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2formed in the reaction.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the mass of CaCl\u2082 formed in this reaction, we first need to establish the limiting reagent and then calculate the amount of CaCl\u2082 produced.<\/p>\n\n\n\n The balanced chemical equation is:<\/p>\n\n\n\n [ The concentration of HCl is 0.76 M (moles per liter), and the volume of HCl is 250 mL (0.250 L). The number of moles of HCl is calculated using the formula:<\/p>\n\n\n\n [ [ The molar mass of CaCO\u2083 is approximately 100.1 g\/mol. The number of moles of CaCO\u2083 is:<\/p>\n\n\n\n [ From the balanced equation, 1 mole of CaCO\u2083 reacts with 2 moles of HCl. Therefore, to fully react with 9.99 moles of CaCO\u2083, we would need:<\/p>\n\n\n\n [ However, we only have 0.19 moles of HCl, which is far less than the 19.98 moles required. Therefore, HCl is the limiting reagent.<\/p>\n\n\n\n From the balanced equation, 2 moles of HCl produce 1 mole of CaCl\u2082. Since HCl is the limiting reagent, we can calculate the moles of CaCl\u2082 formed based on the available moles of HCl:<\/p>\n\n\n\n [ The molar mass of CaCl\u2082 is approximately 147 g\/mol. The mass of CaCl\u2082 produced is:<\/p>\n\n\n\n [ [ This shows that HCl limits the amount of CaCl\u2082 produced, even though there is excess CaCO\u2083.<\/p>\n","protected":false},"excerpt":{"rendered":" Calcium carbonate reacts with aqueous HCl to give CaCl2and CO2according to the reaction given below: CaCO3(s) + 2HCl (aq) ? CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2will be formed when 250 mL of 0.76 M HCI reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2formed […]<\/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-182111","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182111","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=182111"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182111\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182111"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182111"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182111"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Determine the moles of HCl and CaCO\u2083<\/h3>\n\n\n\n
\\text{CaCO\u2083 (s)} + 2\\text{HCl (aq)} \\rightarrow \\text{CaCl\u2082 (aq)} + \\text{CO\u2082 (g)} + \\text{H\u2082O (l)}
]<\/p>\n\n\n\nMoles of HCl:<\/h4>\n\n\n\n
\\text{moles of HCl} = \\text{Molarity (M)} \\times \\text{Volume (L)}
]<\/p>\n\n\n\n
\\text{moles of HCl} = 0.76 \\, \\text{M} \\times 0.250 \\, \\text{L} = 0.19 \\, \\text{moles of HCl}
]<\/p>\n\n\n\nMoles of CaCO\u2083:<\/h4>\n\n\n\n
\\text{moles of CaCO\u2083} = \\frac{\\text{mass (g)}}{\\text{molar mass (g\/mol)}} = \\frac{1000 \\, \\text{g}}{100.1 \\, \\text{g\/mol}} \\approx 9.99 \\, \\text{moles of CaCO\u2083}
]<\/p>\n\n\n\nStep 2: Determine the limiting reagent<\/h3>\n\n\n\n
9.99 \\, \\text{moles of CaCO\u2083} \\times 2 \\, \\text{moles of HCl} = 19.98 \\, \\text{moles of HCl}
]<\/p>\n\n\n\nStep 3: Calculate the moles of CaCl\u2082 produced<\/h3>\n\n\n\n
\\text{moles of CaCl\u2082} = \\frac{0.19 \\, \\text{moles of HCl}}{2} = 0.095 \\, \\text{moles of CaCl\u2082}
]<\/p>\n\n\n\nStep 4: Calculate the mass of CaCl\u2082<\/h3>\n\n\n\n
\\text{mass of CaCl\u2082} = \\text{moles of CaCl\u2082} \\times \\text{molar mass of CaCl\u2082}
]<\/p>\n\n\n\n
\\text{mass of CaCl\u2082} = 0.095 \\, \\text{moles} \\times 147 \\, \\text{g\/mol} = 13.9 \\, \\text{g of CaCl\u2082}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
\n