{"id":257630,"date":"2025-07-17T15:30:08","date_gmt":"2025-07-17T15:30:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257630"},"modified":"2025-07-17T15:30:11","modified_gmt":"2025-07-17T15:30:11","slug":"an-unknown-group-1-metal-carbonate-m2co3-mli-na-or-k-was-reacted-with-excess-2-m-hcl-and-the-mass-of-co2-released-was-determined-by-mass-difference","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/an-unknown-group-1-metal-carbonate-m2co3-mli-na-or-k-was-reacted-with-excess-2-m-hcl-and-the-mass-of-co2-released-was-determined-by-mass-difference\/","title":{"rendered":"An unknown group 1 metal carbonate M2CO3 (M=Li, Na or K) was reacted with excess 2 M HCl and the mass of CO2 released was determined by mass difference"},"content":{"rendered":"\n
An unknown group 1 metal carbonate M2CO3 (M=Li, Na or K) was reacted with excess 2 M HCl and the mass of CO2 released was determined by mass difference. The initial mass of solid M2CO3 was 2.002 g and the mass of CO2 released was 1.206 g. a. Write the balanced chemical equation for the reaction of M2CO3 with HCl. b. What is the mole ratio of CO2 to M2CO3 c. Calculate the molar mass of the unknown metal carbonate and identify the group 1 metal.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let’s break down the problem step by step:<\/p>\n\n\n\n
a) Balanced Chemical Equation<\/h3>\n\n\n\n
When a metal carbonate reacts with hydrochloric acid (HCl), it produces a metal chloride (MCl), carbon dioxide (CO\u2082), and water (H\u2082O). For the unknown metal carbonate M\u2082CO\u2083, the reaction with excess HCl can be written as:M2CO3(s)+2HCl(aq)\u21922MCl(aq)+CO2(g)+H2O(l)M_2CO_3 (s) + 2 HCl (aq) \\rightarrow 2 MCl (aq) + CO_2 (g) + H_2O (l)M2\u200bCO3\u200b(s)+2HCl(aq)\u21922MCl(aq)+CO2\u200b(g)+H2\u200bO(l)<\/p>\n\n\n\n
This equation shows that one mole of M\u2082CO\u2083 reacts with 2 moles of HCl to produce 1 mole of CO\u2082, 2 moles of MCl, and 1 mole of H\u2082O.<\/p>\n\n\n\n
b) Mole Ratio of CO\u2082 to M\u2082CO\u2083<\/h3>\n\n\n\n
From the balanced equation, we can observe that the mole ratio between CO\u2082 and M\u2082CO\u2083 is 1:1. That is, 1 mole of M\u2082CO\u2083 produces 1 mole of CO\u2082.<\/p>\n\n\n\n
c) Calculate the Molar Mass of the Unknown Metal Carbonate and Identify the Group 1 Metal<\/h3>\n\n\n\n\n
Determine the moles of CO\u2082 released:<\/strong><\/li>\n<\/ol>\n\n\n\n
We know the mass of CO\u2082 released (1.206 g). To find the number of moles of CO\u2082, we use the molar mass of CO\u2082, which is 44.01 g\/mol:moles of CO\u2082=1.206\u2009g CO\u208244.01\u2009g\/mol=0.0274\u2009mol CO\u2082\\text{moles of CO\u2082} = \\frac{1.206 \\, \\text{g CO\u2082}}{44.01 \\, \\text{g\/mol}} = 0.0274 \\, \\text{mol CO\u2082}moles of CO\u2082=44.01g\/mol1.206g CO\u2082\u200b=0.0274mol CO\u2082<\/p>\n\n\n\n
\n
Use the mole ratio to find moles of M\u2082CO\u2083:<\/strong><\/li>\n<\/ol>\n\n\n\n
From the balanced equation, we know that the mole ratio of CO\u2082 to M\u2082CO\u2083 is 1:1. Therefore, the moles of M\u2082CO\u2083 are also 0.0274 mol.<\/p>\n\n\n\n
\n
Calculate the molar mass of M\u2082CO\u2083:<\/strong><\/li>\n<\/ol>\n\n\n\n
The mass of M\u2082CO\u2083 used in the reaction is 2.002 g. Using the formula for molar mass:Molar mass of M\u2082CO\u2083=mass of M\u2082CO\u2083moles of M\u2082CO\u2083=2.002\u2009g0.0274\u2009mol=73.1\u2009g\/mol\\text{Molar mass of M\u2082CO\u2083} = \\frac{\\text{mass of M\u2082CO\u2083}}{\\text{moles of M\u2082CO\u2083}} = \\frac{2.002 \\, \\text{g}}{0.0274 \\, \\text{mol}} = 73.1 \\, \\text{g\/mol}Molar mass of M\u2082CO\u2083=moles of M\u2082CO\u2083mass of M\u2082CO\u2083\u200b=0.0274mol2.002g\u200b=73.1g\/mol<\/p>\n\n\n\n
\n
Identify the group 1 metal:<\/strong><\/li>\n<\/ol>\n\n\n\n
The molar mass of M\u2082CO\u2083 is 73.1 g\/mol. The formula for the molar mass of M\u2082CO\u2083 is:M2CO3:2\u00d7(molar mass of M)+60.01\u2009g\/mol (for CO\u2083)M_2CO_3: 2 \\times \\text{(molar mass of M)} + 60.01 \\, \\text{g\/mol (for CO\u2083)}M2\u200bCO3\u200b:2\u00d7(molar mass of M)+60.01g\/mol (for CO\u2083)<\/p>\n\n\n\n
So, the molar mass of M (the metal) can be found by subtracting the molar mass of CO\u2083 (60.01 g\/mol) from the total molar mass:Molar mass of M\u2082CO\u2083=73.1\u2009g\/mol\\text{Molar mass of M\u2082CO\u2083} = 73.1 \\, \\text{g\/mol}Molar mass of M\u2082CO\u2083=73.1g\/mol2\u00d7(molar mass of M)+60.01=73.12 \\times \\text{(molar mass of M)} + 60.01 = 73.12\u00d7(molar mass of M)+60.01=73.12\u00d7(molar mass of M)=73.1\u221260.01=13.092 \\times \\text{(molar mass of M)} = 73.1 – 60.01 = 13.092\u00d7(molar mass of M)=73.1\u221260.01=13.09Molar mass of M=13.092=6.545\u2009g\/mol\\text{Molar mass of M} = \\frac{13.09}{2} = 6.545 \\, \\text{g\/mol}Molar mass of M=213.09\u200b=6.545g\/mol<\/p>\n\n\n\n
The molar mass of the metal is approximately 6.5 g\/mol, which corresponds to Lithium (Li)<\/strong>, with a molar mass of about 6.94 g\/mol.<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
\n
The balanced equation is: M2CO3(s)+2HCl(aq)\u21922MCl(aq)+CO2(g)+H2O(l)M_2CO_3 (s) + 2 HCl (aq) \\rightarrow 2 MCl (aq) + CO_2 (g) + H_2O (l)M2\u200bCO3\u200b(s)+2HCl(aq)\u21922MCl(aq)+CO2\u200b(g)+H2\u200bO(l)<\/li>\n\n\n\n
The mole ratio of CO\u2082 to M\u2082CO\u2083 is 1:1.<\/li>\n\n\n\n
The molar mass of M\u2082CO\u2083 is 73.1 g\/mol, and the metal in the carbonate is Lithium (Li)<\/strong>.<\/li>\n<\/ul>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
An unknown group 1 metal carbonate M2CO3 (M=Li, Na or K) was reacted with excess 2 M HCl and the mass of CO2 released was determined by mass difference. The initial mass of solid M2CO3 was 2.002 g and the mass of CO2 released was 1.206 g. a. Write the balanced chemical equation for 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-257630","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257630","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=257630"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257630\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257630"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257630"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257630"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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