{"id":32309,"date":"2025-06-22T12:44:00","date_gmt":"2025-06-22T12:44:00","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32309"},"modified":"2025-06-22T12:44:02","modified_gmt":"2025-06-22T12:44:02","slug":"mercury-ii-carbonate-will-break-down-when-hgco3-is-exposed-to-heat","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/mercury-ii-carbonate-will-break-down-when-hgco3-is-exposed-to-heat\/","title":{"rendered":"Mercury (II) carbonate will break down when HgCO3 is exposed to heat"},"content":{"rendered":"\n<p>Mercury (II) carbonate will break down when HgCO3 is exposed to heat: HgO + CO2. How much product Hg2O will be formed when 2.50908 g of the reactant is decomposed? (Molar mass of Hg2O is 417.18 g\/mol and molar mass of HgCO3 is 461.19 g\/mol) Select one: a. 1.179 g HgO b. 4.00 g HgO c. 2.269 g HgO d. 0.0977 g HgO<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"486\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-202.png\" alt=\"\" class=\"wp-image-32313\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-202.png 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-202-300x142.png 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-202-768x365.png 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n\n\n\n<p>To solve this problem, we need to use stoichiometry. The reaction provided is:<\/p>\n\n\n\n<p><strong>Hg\u2082CO\u2083 \u2192 Hg\u2082O + CO\u2082<\/strong><\/p>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of Hg\u2082CO\u2083 = 2.50 g<\/li>\n\n\n\n<li>Molar mass (M.m) of Hg\u2082CO\u2083 = 461.19 g\/mol<\/li>\n\n\n\n<li>Molar mass (M.m) of Hg\u2082O = 417.18 g\/mol<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Convert mass of Hg\u2082CO\u2083 to moles<\/h3>\n\n\n\n<p>Moles&nbsp;of&nbsp;Hg\u2082CO\u2083=2.50\u2009g461.19\u2009g\/mol\u22480.00542\u2009mol\\text{Moles of Hg\u2082CO\u2083} = \\frac{2.50 \\, \\text{g}}{461.19 \\, \\text{g\/mol}} \\approx 0.00542 \\, \\text{mol}Moles&nbsp;of&nbsp;Hg\u2082CO\u2083=461.19g\/mol2.50g\u200b\u22480.00542mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the mole ratio from the balanced equation<\/h3>\n\n\n\n<p>From the reaction, <strong>1 mol of Hg\u2082CO\u2083 produces 1 mol of Hg\u2082O<\/strong>. So:Moles&nbsp;of&nbsp;Hg\u2082O=0.00542\u2009mol\\text{Moles of Hg\u2082O} = 0.00542 \\, \\text{mol}Moles&nbsp;of&nbsp;Hg\u2082O=0.00542mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Convert moles of Hg\u2082O to grams<\/h3>\n\n\n\n<p>Mass&nbsp;of&nbsp;Hg\u2082O=0.00542\u2009mol\u00d7417.18\u2009g\/mol\u22482.26\u2009g\\text{Mass of Hg\u2082O} = 0.00542 \\, \\text{mol} \\times 417.18 \\, \\text{g\/mol} \\approx 2.26 \\, \\text{g}Mass&nbsp;of&nbsp;Hg\u2082O=0.00542mol\u00d7417.18g\/mol\u22482.26g<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>c. 2.26 g Hg\u2082O<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>In chemistry, stoichiometry allows us to predict how much of a product will form in a chemical reaction based on the amount of reactant used. In this problem, mercury(I) carbonate (Hg\u2082CO\u2083) decomposes when heated into mercury(I) oxide (Hg\u2082O) and carbon dioxide (CO\u2082). The balanced chemical equation indicates a one-to-one molar relationship between the reactant Hg\u2082CO\u2083 and the product Hg\u2082O.<\/p>\n\n\n\n<p>The first step is to convert the mass of the given reactant into moles using its molar mass. We are given 2.50 grams of Hg\u2082CO\u2083 and its molar mass is 461.19 grams per mole. Dividing these gives us the number of moles of Hg\u2082CO\u2083 involved in the reaction.<\/p>\n\n\n\n<p>Because the molar ratio between Hg\u2082CO\u2083 and Hg\u2082O is one to one, the number of moles of Hg\u2082O formed will be equal to the number of moles of Hg\u2082CO\u2083 decomposed.<\/p>\n\n\n\n<p>Next, we convert the moles of Hg\u2082O into grams using its molar mass, which is 417.18 grams per mole. Multiplying the number of moles of Hg\u2082O by this molar mass gives us the mass of the product formed.<\/p>\n\n\n\n<p>This approach shows how molar relationships and unit conversions can be used to predict the amount of substance produced in a chemical reaction. The correct calculation leads us to find that <strong>2.26 grams of Hg\u2082O<\/strong> will be produced, making option <strong>c<\/strong> the correct answer.<\/p>\n\n\n","protected":false},"excerpt":{"rendered":"<p>Mercury (II) carbonate will break down when HgCO3 is exposed to heat: HgO + CO2. How much product Hg2O will be formed when 2.50908 g of the reactant is decomposed? (Molar mass of Hg2O is 417.18 g\/mol and molar mass of HgCO3 is 461.19 g\/mol) Select one: a. 1.179 g HgO b. 4.00 g HgO [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-32309","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32309","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=32309"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32309\/revisions"}],"predecessor-version":[{"id":32328,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32309\/revisions\/32328"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32309"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32309"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32309"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}