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Answer: For each precipitation reaction, calculate how many grams of the first reactant

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 40P Chapter 8

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 40P

PROBLEM 40P

For each precipitation reaction, calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant.

(a) 2 Kl(aq) + Pb(NO3)2(aq) → Pbl2(s) + 2 KNO3(aq)

(b) Na2CO3(aq) + CuCl2(aq) → CuCO3(s) + 2 NaCl(aq)

(c) K2SO4(aq) + Sr(NO3)2(aq) → SrSO4(s) + 2 KNO3(aq)

Step-by-Step Solution:
Step 1 of 3

Solution 40P

(a) 2 Kl(aq) + Pb(NO3)2(aq) → Pbl2(s) + 2 KNO3(aq)

KI : Pb(NO3)2

2 : 1

So for every one mol of Pb(NO3)2 two mols of KI are required

moles of Pb(NO3)2 = 17.3/331 = 0.052 mol

moles of KI required = 0.052 * 2 = 0.104 mol

Weight of KI = mols * mol wt

= 0.104 * 166

= 17.3 g

(b) Na2CO3(aq) + CuCl2(aq) → CuCO3(s) + 2 NaCl(aq)

Na2CO3 : CuCl2

1 : 1

Moles of CuCl2 = 17.3/135 = 0.128 mol

Moles of Na2CO# required = 0.128 mol

Wt of Na2CO3 = 0.128 * 106 = 13.6 g

(c) K2SO4(aq) + Sr(NO3)2(aq) → SrSO4(s) + 2 KNO3(aq)

K2SO4(aq)+Sr(NO3)2(aq)→SrSO4(s)+2KNO3(aq...

17.3 g of Sr(NO3)2 = 17.3 g/211.63 g/mol = 0.08 mol of Sr(NO3)2

So we need for complete reaction 0.08 mol of K2SO4 i.e. 0.08x174 g = 14.2 g of K2SO4

Step 2 of 3

Chapter 8, Problem 40P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Answer: For each precipitation reaction, calculate how many grams of the first reactant