{"id":25107,"date":"2025-06-18T20:35:05","date_gmt":"2025-06-18T20:35:05","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=25107"},"modified":"2025-06-18T20:35:07","modified_gmt":"2025-06-18T20:35:07","slug":"how-many-moles-of-pbcro4-are-in-50-0-g-of-pbcro4","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-moles-of-pbcro4-are-in-50-0-g-of-pbcro4\/","title":{"rendered":"How many moles of PbCrO4 are in 50.0 g of PbCrO4"},"content":{"rendered":"\n<p>How many moles of PbCrO4 are in 50.0 g of PbCrO4?<\/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>To find how many <strong>moles of PbCrO\u2084<\/strong> (lead(II) chromate) are in <strong>50.0 g<\/strong>, we use the formula: Moles=Mass&nbsp;(g)Molar&nbsp;Mass&nbsp;(g\/mol)\\text{Moles} = \\frac{\\text{Mass (g)}}{\\text{Molar Mass (g\/mol)}}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the molar mass of PbCrO\u2084<\/h3>\n\n\n\n<p>Use the atomic masses from the periodic table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Pb (lead) = 207.2 g\/mol<\/li>\n\n\n\n<li>Cr (chromium) = 52.0 g\/mol<\/li>\n\n\n\n<li>O (oxygen) = 16.0 g\/mol \u00d7 4 = 64.0 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Molar&nbsp;Mass&nbsp;of&nbsp;PbCrO\u2084=207.2+52.0+64.0=323.2&nbsp;g\/mol\\text{Molar Mass of PbCrO\u2084} = 207.2 + 52.0 + 64.0 = 323.2\\ \\text{g\/mol}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the formula to find moles<\/h3>\n\n\n\n<p>Moles&nbsp;of&nbsp;PbCrO\u2084=50.0&nbsp;g323.2&nbsp;g\/mol\u22480.155&nbsp;mol\\text{Moles of PbCrO\u2084} = \\frac{50.0\\ \\text{g}}{323.2\\ \\text{g\/mol}} \\approx 0.155\\ \\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\">\u2705 <strong>Correct Answer:<\/strong><\/h3>\n\n\n\n<p>0.155&nbsp;mol&nbsp;of&nbsp;PbCrO\u2084\\boxed{0.155\\ \\text{mol of PbCrO\u2084}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udd0d Explanation (300 words):<\/h3>\n\n\n\n<p>In chemistry, the number of moles represents the quantity of a substance based on the number of particles (atoms, ions, or molecules) it contains. The mole is a central concept in stoichiometry because it allows chemists to relate mass to the number of particles using the substance\u2019s <strong>molar mass<\/strong> \u2014 the mass of one mole of a compound.<\/p>\n\n\n\n<p>In this problem, we are given 50.0 grams of lead(II) chromate (PbCrO\u2084), and we need to find how many moles this mass represents. The first step is to calculate the <strong>molar mass<\/strong> of PbCrO\u2084 by adding the atomic masses of its elements:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Pb contributes 207.2 g\/mol,<\/li>\n\n\n\n<li>Cr adds 52.0 g\/mol,<\/li>\n\n\n\n<li>and the four oxygen atoms contribute 64.0 g\/mol total.<\/li>\n<\/ul>\n\n\n\n<p>This gives a total molar mass of <strong>323.2 g\/mol<\/strong>.<\/p>\n\n\n\n<p>Next, we apply the basic formula for converting grams to moles: Moles=massmolar&nbsp;mass=50.0323.2\u22480.155\\text{Moles} = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{50.0}{323.2} \\approx 0.155<\/p>\n\n\n\n<p>So, 50.0 grams of PbCrO\u2084 contains approximately <strong>0.155 moles<\/strong>. This calculation is critical in laboratory settings when preparing solutions, conducting reactions, or analyzing sample compositions. Accurate mole calculations ensure proper stoichiometric ratios and successful experimental outcomes.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How many moles of PbCrO4 are in 50.0 g of PbCrO4? The correct answer and explanation is: To find how many moles of PbCrO\u2084 (lead(II) chromate) are in 50.0 g, we use the formula: Moles=Mass&nbsp;(g)Molar&nbsp;Mass&nbsp;(g\/mol)\\text{Moles} = \\frac{\\text{Mass (g)}}{\\text{Molar Mass (g\/mol)}} Step 1: Calculate the molar mass of PbCrO\u2084 Use the atomic masses from the periodic [&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-25107","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25107","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=25107"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25107\/revisions"}],"predecessor-version":[{"id":25108,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25107\/revisions\/25108"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=25107"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=25107"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=25107"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}