# What volume of 0.230 mol/L sodium sulfide solution do I need to react with 30.00 ml of 0.513 mol/L silver nitrate?

Jun 17, 2017

You will need 33.5 mL of $\text{Na"_2"S}$ solution.

#### Explanation:

Step1. Write the balanced chemical equation

$\text{Na"_2"S" + "2AgNO"_3 → "2NaNO"_3 + "Ag"_2"S}$

Step 2. Calculate the moles of ${\text{AgNO}}_{3}$

${\text{Moles of AgNO"_3 = "0.030 00" color(red)(cancel(color(black)("L AgNO"_3))) × "0.513 mol AgNO"_3/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.015 39 mol AgNO}}_{3}$

Step 3. Calculate the moles of $\text{Na"_2"S}$

$\text{Moles of Na"_2"S" = "0.015 39" color(red)(cancel(color(black)("mol AgNO"_3))) × ("1 mol Na"_2"S")/(2 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.007 695 mol Na"_2"S}$

Step 4. Calculate the volume of $\text{Na"_2"S}$

$\text{Vol. of Na"_2"S" = "0.007 695" color(red)(cancel(color(black)("mol Na"_2"S"))) ×("1 L Na"_2"S")/(0.230 color(red)(cancel(color(black)("mol Na"_2"S")))) = "0.0335 L Na"_2"S" = "33.5 mL Na"_2"S}$