The reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) to form silver carbonate (Ag2CO3) and sodium nitrate (NaNO3) is as shown in Equation 1. Roza weighed approximately 2.33 g Na2CO3 (106 g\/mol)<\/p>\n\n\n\n
Na2CO3(aq) + 2AgNO3(aq) Ag2CO3(s) + 2NaNO3(aq) (Equation 1)<\/p>\n\n\n\n
(i) Calculate the mass of AgNO3 (molecular weight = 169.9g\/mol) that is needed to convert 2.33 g of Na2CO3 (molecular weight =106 g\/mol) to Ag2CO3.<\/p>\n\n\n\n
(ii) Calculate the expected yield of Ag2CO3 in g.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine how much AgNO\u2083 is needed to react with 2.33 g of Na\u2082CO\u2083, we first need to use stoichiometry. According to Equation 1:<\/p>\n\n\n\n [ This equation tells us that 1 mole of Na\u2082CO\u2083 reacts with 2 moles of AgNO\u2083. Now, we will follow these steps:<\/p>\n\n\n\n The molecular weight of Na\u2082CO\u2083 is given as 106 g\/mol. To find the moles of Na\u2082CO\u2083 in 2.33 g, use the formula:<\/p>\n\n\n\n [ [ From the balanced equation, we know that 1 mole of Na\u2082CO\u2083 reacts with 2 moles of AgNO\u2083. Therefore:<\/p>\n\n\n\n [ [ Now, we can use the molecular weight of AgNO\u2083 (169.9 g\/mol) to find the mass:<\/p>\n\n\n\n [ [ Thus, 7.47 g of AgNO\u2083<\/strong> is needed to react with 2.33 g of Na\u2082CO\u2083.<\/p>\n\n\n\n To calculate the expected yield of Ag\u2082CO\u2083, we again use stoichiometry. From Equation 1:<\/p>\n\n\n\n [ We already know that 1 mole of Na\u2082CO\u2083 produces 1 mole of Ag\u2082CO\u2083. Therefore, the moles of Ag\u2082CO\u2083 will be the same as the moles of Na\u2082CO\u2083.<\/p>\n\n\n\n From part (i), we calculated that the moles of Na\u2082CO\u2083 in 2.33 g are:<\/p>\n\n\n\n [ From the stoichiometric ratio, the moles of Ag\u2082CO\u2083 formed will be equal to the moles of Na\u2082CO\u2083 used:<\/p>\n\n\n\n [ The molecular weight of Ag\u2082CO\u2083 is:<\/p>\n\n\n\n [ Now, we can calculate the expected mass of Ag\u2082CO\u2083:<\/p>\n\n\n\n [ [ Thus, the expected yield of Ag\u2082CO\u2083 is 6.06 g<\/strong>.<\/p>\n\n\n\n In this problem, we first calculated the amount of silver nitrate (AgNO\u2083) required to completely react with a given amount of sodium carbonate (Na\u2082CO\u2083). Using the stoichiometry from the balanced chemical equation, we found that for 2.33 g of Na\u2082CO\u2083, we needed 7.47 g of AgNO\u2083.<\/p>\n\n\n\n Next, we calculated the expected yield of silver carbonate (Ag\u2082CO\u2083) by considering the moles of Na\u2082CO\u2083 that would react to form Ag\u2082CO\u2083. Since the molar ratio between Na\u2082CO\u2083 and Ag\u2082CO\u2083 is 1:1, the moles of Ag\u2082CO\u2083 formed were the same as the moles of Na\u2082CO\u2083. Using the molar mass of Ag\u2082CO\u2083, we calculated that the expected yield was 6.06 g.<\/p>\n\n\n\n This process involves understanding stoichiometric relationships between reactants and products, and using the molecular weights and molar ratios to convert between grams, moles, and other units.<\/p>\n","protected":false},"excerpt":{"rendered":" The reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) to form silver carbonate (Ag2CO3) and sodium nitrate (NaNO3) is as shown in Equation 1. Roza weighed approximately 2.33 g Na2CO3 (106 g\/mol) Na2CO3(aq) + 2AgNO3(aq) Ag2CO3(s) + 2NaNO3(aq) (Equation 1) (i) Calculate the mass of AgNO3 (molecular weight = 169.9g\/mol) that is needed to […]<\/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-197968","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197968","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=197968"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197968\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197968"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197968"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197968"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(i) Calculation of the mass of AgNO\u2083 needed:<\/strong><\/h3>\n\n\n\n
\\text{Na}_2\\text{CO}_3 (aq) + 2\\text{AgNO}_3 (aq) \\rightarrow \\text{Ag}_2\\text{CO}_3 (s) + 2\\text{NaNO}_3 (aq)
]<\/p>\n\n\n\n\n
\\text{moles of Na}_2\\text{CO}_3 = \\frac{\\text{mass of Na}_2\\text{CO}_3}{\\text{molar mass of Na}_2\\text{CO}_3}
]<\/p>\n\n\n\n
\\text{moles of Na}_2\\text{CO}_3 = \\frac{2.33\\ \\text{g}}{106\\ \\text{g\/mol}} = 0.02197\\ \\text{mol}
]<\/p>\n\n\n\n\n
\\text{moles of AgNO}_3 = 2 \\times \\text{moles of Na}_2\\text{CO}_3
]<\/p>\n\n\n\n
\\text{moles of AgNO}_3 = 2 \\times 0.02197\\ \\text{mol} = 0.04394\\ \\text{mol}
]<\/p>\n\n\n\n\n
\\text{mass of AgNO}_3 = \\text{moles of AgNO}_3 \\times \\text{molar mass of AgNO}_3
]<\/p>\n\n\n\n
\\text{mass of AgNO}_3 = 0.04394\\ \\text{mol} \\times 169.9\\ \\text{g\/mol} = 7.47\\ \\text{g}
]<\/p>\n\n\n\n
\n\n\n\n(ii) Calculation of the expected yield of Ag\u2082CO\u2083:<\/strong><\/h3>\n\n\n\n
\\text{Na}_2\\text{CO}_3 (aq) + 2\\text{AgNO}_3 (aq) \\rightarrow \\text{Ag}_2\\text{CO}_3 (s) + 2\\text{NaNO}_3 (aq)
]<\/p>\n\n\n\n\n
\\text{moles of Na}_2\\text{CO}_3 = 0.02197\\ \\text{mol}
]<\/p>\n\n\n\n\n
\\text{moles of Ag}_2\\text{CO}_3 = 0.02197\\ \\text{mol}
]<\/p>\n\n\n\n\n
\\text{Molar mass of Ag}_2\\text{CO}_3 = 2(107.87) + 12.01 + 3(16.00) = 275.87\\ \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{mass of Ag}_2\\text{CO}_3 = \\text{moles of Ag}_2\\text{CO}_3 \\times \\text{molar mass of Ag}_2\\text{CO}_3
]<\/p>\n\n\n\n
\\text{mass of Ag}_2\\text{CO}_3 = 0.02197\\ \\text{mol} \\times 275.87\\ \\text{g\/mol} = 6.06\\ \\text{g}
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n