Question

How many grams of silver chromate ( Ag2CrO4) will appreciate when 86.0 mL of 0.200 M silver nitrate (AgNO3) is added to 50.0 mL of 0.300 M potassium chromate (K2CrO4)? ( molar mass of Ag2CrO4=331.74 g/mol)

Answer 1

Solution.

$2AgNO3 + K2CrO4 = Ag2CrO4 + 2KNO3$

n(AgNO3) = 0.0172 mol

n(K2CrO4) = 0.0150 mol

If silver nitrate is the limiting agent, the following amount of potassium chromate is required:

$n'(K2CrO4) = \frac{0.0172}{2} = 0.0086 mol$

If potassium chromate is the limiting agent, the following amount of silver nitrate is required:

$n'(AgNO3) = 2 \times 0.0150 = 0.030 mol$

Since in the first case the amount of potassium chromate is sufficient, silver nitrate is the limiting agent.

n(AgNO3) = 2n(Ag2CrO4)

$n(Ag2CrO4) = \frac{n(AgNO3)}{2}$

n(Ag2CrO4) = 0.0086 mol

m(Ag2CrO4) = 2.85 g

Answer:

m(Ag2CrO4) = 2.85 g