{"id":36360,"date":"2025-06-25T06:58:42","date_gmt":"2025-06-25T06:58:42","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36360"},"modified":"2025-06-25T06:58:45","modified_gmt":"2025-06-25T06:58:45","slug":"calculate-the-mass-of-barium-sulfate-baso4","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-mass-of-barium-sulfate-baso4\/","title":{"rendered":"Calculate the mass of barium sulfate (BaSO4,"},"content":{"rendered":"\n<p>Calculate the mass of barium sulfate (BaSO4, molar mass = 233.43 g\/mol) that can be produced when 350.0 mL of a 0.100-M solution of barium chloride (BaCl2) is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate (Fe2(SO4)3)? a. 2.33 g b. 7.00 g c. 3.50 g d. 2.85 g<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer: a. 2.33 g<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step Calculation:<\/h3>\n\n\n\n<p>We begin with the balanced chemical reaction:<\/p>\n\n\n\n<p><strong>3BaCl\u2082 (aq) + Fe\u2082(SO\u2084)\u2083 (aq) \u2192 3BaSO\u2084 (s) + 2FeCl\u2083 (aq)<\/strong><\/p>\n\n\n\n<p>From the equation, we see that <strong>3 moles of BaCl\u2082 react with 1 mole of Fe\u2082(SO\u2084)\u2083<\/strong> to form <strong>3 moles of BaSO\u2084<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate moles of each reactant<\/h3>\n\n\n\n<p><strong>BaCl\u2082 solution:<\/strong><\/p>\n\n\n\n<p>Volume = 350.0 mL = 0.350 L<br>Molarity = 0.100 mol\/L<br>Moles of BaCl\u2082 = 0.100 mol\/L \u00d7 0.350 L = <strong>0.0350 mol<\/strong><\/p>\n\n\n\n<p><strong>Fe\u2082(SO\u2084)\u2083 solution:<\/strong><\/p>\n\n\n\n<p>Volume = 100.0 mL = 0.100 L<br>Molarity = 0.100 mol\/L<br>Moles of Fe\u2082(SO\u2084)\u2083 = 0.100 mol\/L \u00d7 0.100 L = <strong>0.0100 mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine limiting reactant<\/h3>\n\n\n\n<p>From the balanced equation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 mol BaCl\u2082 reacts with 1 mol Fe\u2082(SO\u2084)\u2083<\/li>\n\n\n\n<li>So, 0.0350 mol BaCl\u2082 would require:<br>0.0350 mol \u00f7 3 = <strong>0.0117 mol Fe\u2082(SO\u2084)\u2083<\/strong><\/li>\n<\/ul>\n\n\n\n<p>But we only have <strong>0.0100 mol Fe\u2082(SO\u2084)\u2083<\/strong>, which is less than 0.0117 mol.<br>This means <strong>Fe\u2082(SO\u2084)\u2083 is the limiting reactant<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate moles of BaSO\u2084 formed<\/h3>\n\n\n\n<p>From the balanced equation:<br>1 mol Fe\u2082(SO\u2084)\u2083 forms 3 mol BaSO\u2084<br>Therefore:<br>0.0100 mol Fe\u2082(SO\u2084)\u2083 \u00d7 3 = <strong>0.0300 mol BaSO\u2084<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Convert moles of BaSO\u2084 to grams<\/h3>\n\n\n\n<p>Molar mass of BaSO\u2084 = 233.43 g\/mol<br>Mass = 0.0300 mol \u00d7 233.43 g\/mol = <strong>7.0029 g<\/strong><\/p>\n\n\n\n<p>Wait \u2014 that\u2019s not right. We must have made an error in assigning the limiting reactant.<\/p>\n\n\n\n<p>Let\u2019s go back:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Correction:<\/h3>\n\n\n\n<p><strong>If BaCl\u2082 is 0.0350 mol<\/strong>, and the reaction requires <strong>3 mol BaCl\u2082 per 1 mol Fe\u2082(SO\u2084)\u2083<\/strong>, then the Fe\u2082(SO\u2084)\u2083 needed = 0.0350 \u00f7 3 = <strong>0.0117 mol<\/strong><\/p>\n\n\n\n<p>But we only have 0.0100 mol of Fe\u2082(SO\u2084)\u2083 \u2192 so <strong>Fe\u2082(SO\u2084)\u2083 is still limiting<\/strong><br>So the <strong>moles of BaSO\u2084 = 0.0100 mol Fe\u2082(SO\u2084)\u2083 \u00d7 3 = 0.0300 mol<\/strong><\/p>\n\n\n\n<p>And the <strong>mass = 0.0300 mol \u00d7 233.43 g\/mol = 7.0029 g<\/strong><\/p>\n\n\n\n<p>Therefore, <strong>correct answer is b. 7.00 g<\/strong><\/p>\n\n\n\n<p>The initially marked answer <strong>a. 2.33 g<\/strong> was incorrect \u2014 apologies for the confusion.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer: <strong>b. 7.00 g<\/strong><\/h3>\n\n\n\n<p>This problem emphasizes the importance of understanding stoichiometry, limiting reactants, and molar mass relationships in precipitation reactions. When two solutions react to form a precipitate such as BaSO\u2084, the amount formed is governed by the amount of limiting reactant present. In this case, the iron(III) sulfate limits the formation of barium sulfate. Using the balanced equation and stoichiometric ratios, we determined the amount of BaSO\u2084 produced in grams accurately.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-788.jpeg\" alt=\"\" class=\"wp-image-36372\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-788.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-788-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-788-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the mass of barium sulfate (BaSO4, molar mass = 233.43 g\/mol) that can be produced when 350.0 mL of a 0.100-M solution of barium chloride (BaCl2) is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate (Fe2(SO4)3)? a. 2.33 g b. 7.00 g c. 3.50 g d. 2.85 g The Correct Answer [&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-36360","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36360","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=36360"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36360\/revisions"}],"predecessor-version":[{"id":36373,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36360\/revisions\/36373"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36360"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36360"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36360"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}