Questions: According to the following reaction, what volume of 0.305 M AgNO3 is required to react with 155.0 mL of 0.274 M Na2SO4 solution? 2 AgNO3(aq) + Na2SO4(aq) → Ag2SO 4(s) + 2 NaNO3(aq) 139 mL 345 mL 278 mL 173 mL 581 mL

Solution Steps

Calculate the moles of $\text{Na}_{2} \text{SO}_{4}$ using its concentration and volume: $$\text{Moles of } \text{Na}_{2} \text{SO}_{4} = 0.274 \, \text{M} \times 0.155 \, \text{L} = 0.04247 \, \text{mol}$$

From the balanced chemical equation, the stoichiometric ratio between $\text{AgNO}_{3}$ and $\text{Na}_{2} \text{SO}_{4}$ is 2:1. Therefore, calculate the moles of $\text{AgNO}_{3}$ needed: $$\text{Moles of } \text{AgNO}_{3} = 2 \times 0.04247 \, \text{mol} = 0.08494 \, \text{mol}$$

Use the concentration of $\text{AgNO}_{3}$ to find the required volume: $$\text{Volume of } \text{AgNO}_{3} = \frac{0.08494 \, \text{mol}}{0.305 \, \text{M}} = 0.278 \, \text{L} = 278 \, \text{mL}$$

Final Answer