Question #195272

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

Expert's answer

The reaction equation:

2AgNO_{3} + K_{2}CrO_{4} = Ag_{2}CrO_{4} + 2KNO_{3}

Quantity of AgNO_{3}:

0.086 L x 0.2 M = 0.0172 mol

Quantity of K_{2}CrO_{4}:

0.50 L x 6.3 M = 3.15 mol

According to the reaction equation each 2 moles of AgNO_{3} reacts with 1 mole of K_{2}CrO_{4}.

So, K_{2}CrO_{4} is in excess.

Therefore, the quantity of Ar_{2}CrO_{4} = 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

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