Question

How many grams of potassium nitrate (KNO3) would form if 2.25 liters of a 1.50 molar lead nitrate Pb(NO3)2 solution reacts with 1.15 liters of a 2.75 molar potassium chromate K2CrO4 solution?

Pb(NO3)2 + K2CrO4 PbCrO4 + 2KNO3

108 g
681 g
639 g
362 g
@nechirwan

Answer 1

Balanced equation:
Pb(NO3)2 + K2CrO4 → PbCrO4 + 2KNO3
1mol Pb(NO3)2 reacts with 1mol K2CrO4 to produce 2 mol KNO3
Check for limiting reactant
Mol Pb(NO3)2 in 2.25L of 1.5M solution = 2.25*1.5 = 3.375 mol
Mol K2CrO4 in 1.15L of 2.75M solution = 1.15*2.75 = 3.1625 mol

The K2CrO4 is limiting
From the equation: 1mol K2CrO4 will produce 2 mol KNO3
3.1625 mol K2CrO4 will produce 3.1625*2 = 6.325 mol KNO3
Molar mass KNO3 = 101.1g/mol
6.325 mol = 6.325*101.1 = 639 g