How many grams of Ag2 Cr04 will precipitate when 86ml of 0.2 molar Ag2 Cr04 is added to 500ml
of 6.3 molar K2CrO4? Molar mass of Ag2CrO4 = 331.74 g/mol.
Balanced Equation: 2AgNO3 + K2CrO4 Ag2 Cr04 + 2KN03
The reaction equation:
2AgNO3 + K2CrO4 = Ag2CrO4 + 2KNO3
Quantity of AgNO3:
0.086 L x 0.2 M = 0.0172 mol
Quantity of K2CrO4:
0.50 L x 6.3 M = 3.15 mol
According to the reaction equation each 2 moles of AgNO3 reacts with 1 mole of K2CrO4.
So, K2CrO4 is in excess.
Therefore, the quantity of Ar2CrO4 = 0.0172 / 2 = 0.0086 mol (according to the equation).
0.0086 mol x 331.74 g/mol = 2.853 g
Answer: 2.853 g