Answer to Question #195272 in General Chemistry for Jerick

Question

 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

Answer 1

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).

The mass:

0.0086 mol x 331.74 g/mol = 2.853 g

Answer: 2.853 g