100 m 0125 l 125 102 mol 1 mol pbno3 2 331 g pbno3 2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: formed. (b) Calculate the number of moles of each reactant. mol Pb(NO3)2 = 0.150 g Pb(NO3)2 × = 4.53 × 10−4 mol mol NaI = 0.100 M × 0.125 L = 1.25 × 10−2 mol 1 mol Pb(NO3 )2 331 g Pb(NO3 )2 One point is earned for the correct number of moles of Pb(NO3)2. One point is earned for the correct number of moles of NaI. (c) Identify the limiting reactant. Show calculations to support your identification. 2 mol NaI 1 mol Pb(NO3 )2 mol NaI reacting = 4.53 × 10−4 mol Pb(NO3)2 × = 9.06 × 10−4 mol One point is earned for the identification of Pb(NO3)2. One point is earned for a justification in terms of the relative numbers of moles. There is 1.25 × 10−2 mol of NaI initially, thus Pb(NO3)2 is the limiting reactant. (d) Calculate the molar concentration of NO3−(aq) in the mixture after the reaction is complete. 2 × (4.53 × 10−4 mol) = = 7.25 × 10−3 M f 0.125 L [NO3 −] One point is earned for the correct NO3−/Pb2+ stoichiometry. One point is earned for the correct molarity. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® CHEMISTRY 200...
View Full Document

This note was uploaded on 05/05/2011 for the course CHEM 504 taught by Professor John during the Fall '11 term at American College of Computer & Information Sciences.

Ask a homework question - tutors are online