Let’s break this down step by step.<\/p>\n\n\n\n
Part A: Writing the Net Ionic Equation<\/h3>\n\n\n\n
1. Write the balanced molecular equation:<\/strong><\/p>\n\n\n\n The reaction between potassium chloride (KCl) and lead (II) nitrate (Pb(NO\u2083)\u2082) forms lead (II) chloride (PbCl\u2082) as a precipitate and potassium nitrate (KNO\u2083) in solution:<\/p>\n\n\n\n Pb(NO\u2083)\u2082 (aq)+2KCl (aq)\u2192PbCl\u2082 (s)+2KNO\u2083 (aq)\\text{Pb(NO\u2083)\u2082 (aq)} + 2\\text{KCl (aq)} \\rightarrow \\text{PbCl\u2082 (s)} + 2\\text{KNO\u2083 (aq)}Pb(NO\u2083)\u2082 (aq)+2KCl (aq)\u2192PbCl\u2082 (s)+2KNO\u2083 (aq)<\/p>\n\n\n\n 2. Dissociate the aqueous compounds into ions:<\/strong><\/p>\n\n\n\n The net ionic equation will involve only the ions that participate in the reaction.<\/p>\n\n\n\n 3. Eliminate spectator ions:<\/strong><\/p>\n\n\n\n Spectator ions are ions that do not change during the reaction. In this case, the potassium ion (K\u207a) and the nitrate ion (NO\u2083\u207b) are spectator ions. Therefore, they are removed from the net ionic equation.<\/p>\n\n\n\n Net ionic equation:<\/strong>Pb\u00b2\u207a (aq)+2Cl\u207b (aq)\u2192PbCl\u2082 (s)\\text{Pb\u00b2\u207a (aq)} + 2\\text{Cl\u207b (aq)} \\rightarrow \\text{PbCl\u2082 (s)}Pb\u00b2\u207a (aq)+2Cl\u207b (aq)\u2192PbCl\u2082 (s)<\/p>\n\n\n\n This shows the formation of lead (II) chloride as a precipitate.<\/p>\n\n\n\n 1. Moles of potassium chloride (KCl):<\/strong><\/p>\n\n\n\n We are given 150 mL of 0.583 M KCl. First, calculate the moles of KCl:moles of KCl=0.150\u2009L\u00d70.583\u2009mol\/L=0.08745\u2009mol of KCl\\text{moles of KCl} = 0.150 \\, \\text{L} \\times 0.583 \\, \\text{mol\/L} = 0.08745 \\, \\text{mol of KCl}moles of KCl=0.150L\u00d70.583mol\/L=0.08745mol of KCl<\/p>\n\n\n\n Since the stoichiometric ratio between KCl and PbCl\u2082 is 2:1, the moles of PbCl\u2082 formed will be half of the moles of KCl:moles of PbCl\u2082=0.08745\u2009mol of KCl2=0.043725\u2009mol of PbCl\u2082\\text{moles of PbCl\u2082} = \\frac{0.08745 \\, \\text{mol of KCl}}{2} = 0.043725 \\, \\text{mol of PbCl\u2082}moles of PbCl\u2082=20.08745mol of KCl\u200b=0.043725mol of PbCl\u2082<\/p>\n\n\n\n 2. Calculate the number of molecules of PbCl\u2082:<\/strong><\/p>\n\n\n\n Now, use Avogadro\u2019s number (6.022 \u00d7 10\u00b2\u00b3 molecules\/mol) to find the number of molecules of PbCl\u2082:molecules of PbCl\u2082=0.043725\u2009mol of PbCl\u2082\u00d76.022\u00d71023\u2009molecules\/mol\\text{molecules of PbCl\u2082} = 0.043725 \\, \\text{mol of PbCl\u2082} \\times 6.022 \\times 10^{23} \\, \\text{molecules\/mol}molecules of PbCl\u2082=0.043725mol of PbCl\u2082\u00d76.022\u00d71023molecules\/molmolecules of PbCl\u2082=2.63\u00d71022\u2009molecules of PbCl\u2082\\text{molecules of PbCl\u2082} = 2.63 \\times 10^{22} \\, \\text{molecules of PbCl\u2082}molecules of PbCl\u2082=2.63\u00d71022molecules of PbCl\u2082<\/p>\n\n\n\n 150 mL of 0.583 M potassium chloride (74.55 g\/mol) reacts with an excess of lead (II) nitrate (325.39 g\/mol). Write the net ionic equation for this reaction. Clearly indicate charges and states of matter as appropriate. b: How many molecules of precipitate are theoretically formed in this reaction? The Correct Answer and Explanation is: Let’s […]<\/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-42663","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42663","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=42663"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42663\/revisions"}],"predecessor-version":[{"id":42665,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42663\/revisions\/42665"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42663"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42663"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42663"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Part B: Calculating the Number of Molecules of Precipitate<\/h3>\n\n\n\n
Summary:<\/h3>\n\n\n\n
\n
Pb\u00b2\u207a\u00a0(aq)+2Cl\u207b\u00a0(aq)\u2192PbCl\u2082\u00a0(s)\\text{Pb\u00b2\u207a (aq)} + 2\\text{Cl\u207b (aq)} \\rightarrow \\text{PbCl\u2082 (s)}Pb\u00b2\u207a\u00a0(aq)+2Cl\u207b\u00a0(aq)\u2192PbCl\u2082\u00a0(s)<\/li>\n\n\n\n
2.63\u00d71022\u2009molecules2.63 \\times 10^{22} \\, \\text{molecules}2.63\u00d71022molecules<\/li>\n<\/ul>\n\n\n\n![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-351.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"