Question

How many grams of silver chromate will precipitate when 250.0 mL of 0.350 M silver nitrate are added to 250.0 mL of 0.450 M potassium chromate?

Answer 1

2 AgNO3(aq) + K2CrO4(aq)  Ag2CrO4(s) + 2 KNO3(aq)

Step-by-step explanation:

0.150 L AgNO3 0.500 moles AgNO3 1 moles Ag2CrO4 331.74 g Ag2CrO4 = 12.4 g Ag2CrO4

1 L 2 moles AgNO3 1 moles Ag2CrO4

0.100 L K2CrO4 0.400 moles K2CrO4 1 moles Ag2CrO4 331.74 g Ag2CrO4 = 13.3 g Ag2CrO4

1 L 1 moles K2CrO4 1 moles Ag2CrO4