How many millimeters of .230 M Na2S are needed to react with 30.00 mL of .513 M AgNO3, according to the following balanced equation? Na2S (aq) + 2AgNO3 (aq) ---> 2NaNO3 (aq) + Ag2S (s)

1 Answer
Jun 17, 2017

Answer:

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"#