Question

If you mix solutions containing 0.014200 mols of potassium chromate with 0.00190 mols of silver nitrate, then what is the maximum mass in grams (g) of Ag2CrO4(s) that can be produceu?

Answer 1

Precipitation reaction;

K2CrO4(aq) + 2AgNO3(aq) >>>>>>> Ag2CrO4(s) + 2KNO3(aq)

1 mol of pot.chromate reacts with 2 moles AgNO3 completely to form 1 mol Ag2CrO4(s) .

Available moles of AgNO3 (0.00190) are less stoichiometrically for complete reaction of moles of pot.chromate given (0.014200) .

For 0.014200 mols pot.chromate , we need = (2mol.AgNO3/1mol-pot.chromate)×0.014200mol-pot.chromate = 0.0284 mol -AgNO3 required for complete reaction.

Thus AgNO3 is the limiting reagent and amounts of product formed depends on the amount of limiting reagent available.

0.00190 mol AgNO3 will produce = (1mol - Ag2CrO4/2mol-AgNO3)×0.00190mol-AgNO3 = 0.00095 mol Ag2CrO4 .

Mass of Ag2CrO4 will form = moles × molar mass = 0.00095moles × 331.73g/mol = 0.315 g of Ag2CrO4.

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