Answer to Question #139248 in General Chemistry for Jane

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

The reaction equation:

2AgNO₃ + K₂CrO₄ = Ag₂CrO₄ + 2KNO₃

Quantity of AgNO₃:

0.086 L x 0.200 M = 0.0172 mol

Quantity of K₂CrO₄:

0.050 L x 0.300 M = 0.0150 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