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Chapter%2011_Examples

Chapter%2011_Examples - CHM 11200 Spring 10 Chapter 11...

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CHM 11200 Spring 10 Chapter 11 Worked Lecture Examples
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Example 1 (Practice Problem 11.8, p 441) 2 When a solution of lead(II) nitrate is mixed with a solution of potassium chromate, a yellow precipitate forms according to the equation Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2 KNO 3 (aq) What volume of 0.105 M lead(II) nitrate is required to react with 100.0 mL of 0.120 M potassium chromate? What mass of PbCrO 4 solid forms? First, calculate the number of moles of K 2 CrO 4 : 0.1000 L ( ) 0.120 mol 1 L = 0.0120 mol K 2 CrO 4
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Example 1 (cont’d) 3 First, calculate the number of moles of K 2 CrO 4 : 0.1000 L ( ) 0.120 mol 1 L = 0.0120 mol K 2 CrO 4 Next, calculate the number of moles of Pb(NO 3 ) 2 needed to react with the 0.0120 mol K 2 CrO 4 : mol Pb(NO 3 ) 2 = 0.0120 mol K 2 CrO 4 ( ) 1 mol Pb(NO 3 ) 2 1 mol K 2 CrO 4 = 0.0120 mol Pb(NO 3 ) 2 Next, calculate the volume of 0.105 M Pb(NO 3 ) 2 needed: 0.0120 mol Pb(NO 3 ) 2 1 L 0.105 mol = 0.114 L or 114 mL
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Example 1 (cont’d) 4 Now, calculate the number of moles of PbCrO 4 formed:
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