# Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 181: 34

2.4 Liters

#### Work Step by Step

Moles of $AgNO_3$ = $10. \text{ grams}*\frac{1 \text{ mol}}{169.87 \text{ grams}}=0.059$ moles $0.59 \text{ moles}*\frac{1 \text{ L}}{0.25 \text{ moles}}=2.4$ L

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