Respuesta :
The amount of silver chromate that precipitates after addition of solutions is 12.44 g.
Number of moles:
The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:
Moles = Molarity x Volume
Calculations:
Step 1:
The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:
Moles of AgNO3 = 0.500 M x (150/1000) L
= 0.075 mol
Step 2:
The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:
Moles of K2CrO4 = 0.400 M x (100/1000) L
= 0.04 mol
Step 3:
The balanced chemical reaction between AgNO3 and K2CrO4 is:
2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3
The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol
The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.
Step 4:
According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.
The molar mass of Ag2CrO4 is 331.74 g/mol.
The mass of Ag2CrO4 is calculated as:
Mass = 0.0375 mol x 331.74 g/mol
= 12.44 g
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