To determine the mass of aluminium nitrate that was originally in the solution, we can start by considering the stoichiometry of the reaction given:<\/p>\n\n\n\n
[ 2 \\text{Al(NO}_3)_3 (aq) + 3 \\text{Na}_2\\text{CO}_3 (aq) \\rightarrow \\text{Al}_2(\\text{CO}_3)_3 (s) + 6 \\text{NaNO}_3 (aq) ]<\/p>\n\n\n\n
From the reaction, you can see that 2 moles of aluminium nitrate react to produce 1 mole of aluminium carbonate. First, let\u2019s calculate the molar mass of aluminium carbonate (Al\u2082(CO\u2083)\u2083):<\/p>\n\n\n\n
- \n
- Molar mass of Al: 26.98 g\/mol<\/li>\n\n\n\n
- Molar mass of C: 12.01 g\/mol<\/li>\n\n\n\n
- Molar mass of O: 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n
The molar mass of Al\u2082(CO\u2083)\u2083 = ( 2 \\times 26.98 + 3 \\times (12.01 + 3 \\times 16.00) ) g\/mol = ( 53.96 + 3 \\times 60.01 ) g\/mol = 233.99 g\/mol.<\/p>\n\n\n\n
Given that 4.68 g of Al\u2082(CO\u2083)\u2083 is formed, the number of moles of Al\u2082(CO\u2083)\u2083 can be calculated as:<\/p>\n\n\n\n
[ \\text{Moles of Al}_2(\\text{CO}_3)_3 = \\frac{4.68 \\text{ g}}{233.99 \\text{ g\/mol}} \\approx 0.020 \\text{ mol} ]<\/p>\n\n\n\n
Using the stoichiometry of the reaction (2:1 ratio between Al(NO\u2083)\u2083 and Al\u2082(CO\u2083)\u2083), the moles of Al(NO\u2083)\u2083 required would be double the moles of Al\u2082(CO\u2083)\u2083, which is ( 0.020 \\times 2 = 0.040 \\text{ mol} ).<\/p>\n\n\n\n
Next, calculate the molar mass of Al(NO\u2083)\u2083:<\/p>\n\n\n\n
- \n
- Molar mass of Al(NO\u2083)\u2083 = 26.98 + 3 \\times (14.01 + 3 \\times 16.00) g\/mol = 26.98 + 3 \\times 62.01 g\/mol = 212.98 g\/mol.<\/li>\n<\/ul>\n\n\n\n
Thus, the mass of Al(NO\u2083)\u2083 is:
[ \\text{Mass of Al(NO}_3)_3 = 0.040 \\text{ mol} \\times 212.98 \\text{ g\/mol} = 8.52 \\text{ g} ]<\/p>\n\n\n\nHence, 8.52 g of aluminium nitrate was originally in the solution.<\/p>\n\n\n\n
The calculation shows how chemical stoichiometry helps in relating the masses of reactants and products in a chemical reaction. By understanding the mole ratio between reactants and products, we can precisely determine the amount of reactants needed to form a certain amount of product and vice versa. This is fundamental in fields such as chemical engineering, pharmaceuticals, and any scientific research involving chemical reactions. This demonstrates the importance of balanced chemical equations and molar mass in practical applications like calculating reactant quantities in laboratory or industrial processes.<\/p>\n\n\n\n
Now, let’s generate an image to visually represent this chemical reaction:<\/p>\n\n\n\n
- \n
- The image will include labelled reactant and product bottles with chemical formulas, a reaction arrow showing the conversion from reactants to products, and scales showing the calculated masses.<\/li>\n<\/ul>\n\n\n\n
Let me generate the image for you.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1442.png\")
<\/figure>\n\n\n\nHere is the visual representation of the chemical reaction between aluminium nitrate and sodium carbonate, illustrating the formation of aluminium carbonate and sodium nitrate, complete with labelled reactants and products, as well as the calculated masses on a balance scale. This image provides a clear view of the laboratory setup for this reaction.<\/p>\n","protected":false},"excerpt":{"rendered":"
The mass of aluminium nitrate in a solution is determined by adding sodium carbonate solution to form aluminium carbonate and sodium nitrate. The equation for the reaction occurring is: 2Al(NO3)3(aq) + 3Na2CO3(aq) \u2192 Al2(CO3)3(s) + 6NaNO3(aq) In a particular reaction, 4.68g of aluminium carbonate is obtained. The Correct Answer and Explanation is : To determine […]<\/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-194806","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194806","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=194806"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194806\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194806"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194806"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194806"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- The image will include labelled reactant and product bottles with chemical formulas, a reaction arrow showing the conversion from reactants to products, and scales showing the calculated masses.<\/li>\n<\/ul>\n\n\n\n
- Molar mass of Al(NO\u2083)\u2083 = 26.98 + 3 \\times (14.01 + 3 \\times 16.00) g\/mol = 26.98 + 3 \\times 62.01 g\/mol = 212.98 g\/mol.<\/li>\n<\/ul>\n\n\n\n