Question

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?

Answer 1

You will need 33.5 mL of `"Na"_2"S""` solution.

Explanation:

Step1. Write the balanced chemical equation

`"Na"_2"S" + "2AgNO"_3 → "2NaNO"_3 + "Ag"_2"S"`

Step 2. Calculate the moles of `"AgNO"_3`

`"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 `"Na"_2"S"`

`"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 `"Na"_2"S"`

`"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"`