{"id":20199,"date":"2025-06-14T08:20:59","date_gmt":"2025-06-14T08:20:59","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=20199"},"modified":"2025-06-14T08:21:07","modified_gmt":"2025-06-14T08:21:07","slug":"a-precipitation-reaction-occurs-when-50-0-ml-of-0-50-m-bacl2aq-is-mixed-with-75-0-ml-of-0-75-m-na2co3aq","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-precipitation-reaction-occurs-when-50-0-ml-of-0-50-m-bacl2aq-is-mixed-with-75-0-ml-of-0-75-m-na2co3aq\/","title":{"rendered":"A precipitation reaction occurs when 50.0 mL of 0.50 M BaCl2(aq) is mixed with 75.0 mL of 0.75 M Na2CO3(aq)"},"content":{"rendered":"\n<p>A precipitation reaction occurs when 50.0 mL of 0.50 M BaCl2(aq) is mixed with 75.0 mL of 0.75 M Na2CO3(aq). The only precipitate is the BaCO3(s) formed. Which is the limiting reactant?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s analyze the precipitation reaction step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of BaCl\u2082 = 50.0 mL = 0.0500 L<\/li>\n\n\n\n<li>Molarity of BaCl\u2082 = 0.50 M<\/li>\n\n\n\n<li>Volume of Na\u2082CO\u2083 = 75.0 mL = 0.0750 L<\/li>\n\n\n\n<li>Molarity of Na\u2082CO\u2083 = 0.75 M<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Write the balanced chemical equation<\/h3>\n\n\n\n<p>The reaction between barium chloride and sodium carbonate produces barium carbonate precipitate and sodium chloride in solution: BaCl2(aq)+Na2CO3(aq)\u2192BaCO3(s)+2NaCl(aq)\\text{BaCl}_2(aq) + \\text{Na}_2\\text{CO}_3(aq) \\rightarrow \\text{BaCO}_3(s) + 2\\text{NaCl}(aq)<\/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: Calculate moles of each reactant<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of BaCl\u2082:<\/li>\n<\/ul>\n\n\n\n<p>moles&nbsp;BaCl2=M\u00d7V=0.50\u2009mol\/L\u00d70.0500\u2009L=0.025\u2009mol\\text{moles BaCl}_2 = M \\times V = 0.50 \\, \\text{mol\/L} \\times 0.0500 \\, \\text{L} = 0.025 \\, \\text{mol}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of Na\u2082CO\u2083:<\/li>\n<\/ul>\n\n\n\n<p>moles&nbsp;Na2CO3=0.75\u2009mol\/L\u00d70.0750\u2009L=0.05625\u2009mol\\text{moles Na}_2\\text{CO}_3 = 0.75 \\, \\text{mol\/L} \\times 0.0750 \\, \\text{L} = 0.05625 \\, \\text{mol}<\/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: Determine the limiting reactant<\/h3>\n\n\n\n<p>From the balanced equation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mole of BaCl\u2082 reacts with 1 mole of Na\u2082CO\u2083 to produce 1 mole of BaCO\u2083.<\/li>\n<\/ul>\n\n\n\n<p>So, the mole ratio is 1:1.<\/p>\n\n\n\n<p>Compare moles available:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>BaCl\u2082: 0.025 mol<\/li>\n\n\n\n<li>Na\u2082CO\u2083: 0.05625 mol<\/li>\n<\/ul>\n\n\n\n<p>Since BaCl\u2082 has fewer moles, it will run out first.<\/p>\n\n\n\n<p><strong>Therefore, BaCl\u2082 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\">Explanation (300 words):<\/h3>\n\n\n\n<p>In a precipitation reaction involving BaCl\u2082 and Na\u2082CO\u2083, the formation of the solid BaCO\u2083 depends on the availability of the reactants. To identify the limiting reactant\u2014the substance that is completely consumed first and thus limits the amount of product formed\u2014we first write the balanced chemical equation: BaCl2+Na2CO3\u2192BaCO3(s)+2NaCl\\text{BaCl}_2 + \\text{Na}_2\\text{CO}_3 \\rightarrow \\text{BaCO}_3(s) + 2\\text{NaCl}<\/p>\n\n\n\n<p>The stoichiometry indicates that one mole of barium chloride reacts with one mole of sodium carbonate to produce one mole of barium carbonate precipitate.<\/p>\n\n\n\n<p>Next, we calculate the number of moles of each reactant:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For BaCl\u2082, with 0.50 M concentration and 50.0 mL volume, the moles are:<\/li>\n<\/ul>\n\n\n\n<p>0.50\u2009mol\/L\u00d70.0500\u2009L=0.025\u2009mol0.50 \\, \\text{mol\/L} \\times 0.0500 \\, \\text{L} = 0.025 \\, \\text{mol}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For Na\u2082CO\u2083, with 0.75 M concentration and 75.0 mL volume, the moles are:<\/li>\n<\/ul>\n\n\n\n<p>0.75\u2009mol\/L\u00d70.0750\u2009L=0.05625\u2009mol0.75 \\, \\text{mol\/L} \\times 0.0750 \\, \\text{L} = 0.05625 \\, \\text{mol}<\/p>\n\n\n\n<p>Since the reaction requires a 1:1 mole ratio, we compare the available moles of each reactant. BaCl\u2082 has 0.025 moles available, whereas Na\u2082CO\u2083 has more, at 0.05625 moles. This means BaCl\u2082 will be consumed first during the reaction because there is less of it in terms of moles.<\/p>\n\n\n\n<p>Hence, <strong>BaCl\u2082 is the limiting reactant<\/strong>. Once all BaCl\u2082 is used up, the reaction will stop, regardless of the remaining Na\u2082CO\u2083. The amount of BaCO\u2083 precipitate formed will correspond to the amount of BaCl\u2082 initially present.<\/p>\n\n\n\n<p>Understanding the limiting reactant helps in predicting the theoretical yield of a product and ensures efficient use of chemicals in laboratory and industrial settings.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A precipitation reaction occurs when 50.0 mL of 0.50 M BaCl2(aq) is mixed with 75.0 mL of 0.75 M Na2CO3(aq). The only precipitate is the BaCO3(s) formed. Which is the limiting reactant? The correct answer and explanation is: Let&#8217;s analyze the precipitation reaction step-by-step. Given: Step 1: Write the balanced chemical equation The reaction between [&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-20199","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20199","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=20199"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20199\/revisions"}],"predecessor-version":[{"id":20200,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20199\/revisions\/20200"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=20199"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=20199"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=20199"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}