Answer to Question #195272 in General Chemistry for Jerick

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


1
Expert's answer
2021-05-21T02:32:59-0400

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



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