{"id":41434,"date":"2025-06-28T08:40:28","date_gmt":"2025-06-28T08:40:28","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=41434"},"modified":"2025-06-28T08:40:29","modified_gmt":"2025-06-28T08:40:29","slug":"calculate-the-mass-of-pbi2-produced-by-treatment-of-0-995-g-of-ki-with-excess-pbno32","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-mass-of-pbi2-produced-by-treatment-of-0-995-g-of-ki-with-excess-pbno32\/","title":{"rendered":"Calculate the mass of PbI2 produced by treatment of 0.995 g of KI with excess Pb(NO3)2"},"content":{"rendered":"\n<p>Calculate the mass of PbI2 produced by treatment of 0.995 g of KI with excess Pb(NO3)2: a. 1.38 g b. 2.76 g c. 11.1 g d. 1.00 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>To find the mass of PbI\u2082 produced when 0.995 g of KI reacts with excess Pb(NO\u2083)\u2082, follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 1: Write the balanced chemical equation<\/strong><\/p>\n\n\n\n<p>2KI+Pb(NO3)2\u2192PbI2+2KNO32KI + Pb(NO_3)_2 \\rightarrow PbI_2 + 2KNO_32KI+Pb(NO3\u200b)2\u200b\u2192PbI2\u200b+2KNO3\u200b<\/p>\n\n\n\n<p>This tells us that 2 moles of potassium iodide (KI) react with 1 mole of lead(II) nitrate to form 1 mole of lead(II) iodide (PbI\u2082).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 2: Calculate the molar mass of KI<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Potassium (K): 39.10 g\/mol<\/li>\n\n\n\n<li>Iodine (I): 126.90 g\/mol<\/li>\n\n\n\n<li>Molar mass of KI = 39.10 + 126.90 = <strong>166.00 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Now, calculate moles of KI:Moles&nbsp;of&nbsp;KI=0.995&nbsp;g166.00&nbsp;g\/mol=0.00599&nbsp;mol\\text{Moles of KI} = \\frac{0.995\\ \\text{g}}{166.00\\ \\text{g\/mol}} = 0.00599\\ \\text{mol}Moles&nbsp;of&nbsp;KI=166.00&nbsp;g\/mol0.995&nbsp;g\u200b=0.00599&nbsp;mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 3: Use stoichiometry to find moles of PbI\u2082 formed<\/strong><\/p>\n\n\n\n<p>From the balanced equation:2&nbsp;mol&nbsp;KI\u21921&nbsp;mol&nbsp;PbI22\\ \\text{mol KI} \\rightarrow 1\\ \\text{mol PbI}_22&nbsp;mol&nbsp;KI\u21921&nbsp;mol&nbsp;PbI2\u200bMoles&nbsp;of&nbsp;PbI2=0.005992=0.002995&nbsp;mol\\text{Moles of PbI}_2 = \\frac{0.00599}{2} = 0.002995\\ \\text{mol}Moles&nbsp;of&nbsp;PbI2\u200b=20.00599\u200b=0.002995&nbsp;mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 4: Calculate the mass of PbI\u2082<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Lead (Pb): 207.2 g\/mol<\/li>\n\n\n\n<li>Iodine (I\u2082): 2 \u00d7 126.9 = 253.8 g\/mol<\/li>\n\n\n\n<li>Molar mass of PbI\u2082 = 207.2 + 253.8 = <strong>461.0 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Mass&nbsp;of&nbsp;PbI2=0.002995&nbsp;mol\u00d7461.0&nbsp;g\/mol=1.38&nbsp;g\\text{Mass of PbI}_2 = 0.002995\\ \\text{mol} \\times 461.0\\ \\text{g\/mol} = 1.38\\ \\text{g}Mass&nbsp;of&nbsp;PbI2\u200b=0.002995&nbsp;mol\u00d7461.0&nbsp;g\/mol=1.38&nbsp;g<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Correct answer: a. 1.38 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>In this reaction, KI is the limiting reagent, and Pb(NO\u2083)\u2082 is in excess. We used the balanced equation to determine the mole ratio. Then we converted grams of KI to moles, applied the stoichiometry to find the moles of PbI\u2082, and converted that to grams. This approach ensures the correct theoretical yield is calculated based on limiting reactant principles and molar mass relationships.<\/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-banner8-1336.jpeg\" alt=\"\" class=\"wp-image-41444\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1336.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1336-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1336-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the mass of PbI2 produced by treatment of 0.995 g of KI with excess Pb(NO3)2: a. 1.38 g b. 2.76 g c. 11.1 g d. 1.00 g The Correct Answer and Explanation is: To find the mass of PbI\u2082 produced when 0.995 g of KI reacts with excess Pb(NO\u2083)\u2082, follow these steps: Step 1: [&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-41434","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41434","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=41434"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41434\/revisions"}],"predecessor-version":[{"id":41445,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41434\/revisions\/41445"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41434"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41434"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}