Answer to Question #319199 in General Chemistry for Emman

Question

When 200 mL of 0.700 molarity silver nitrate is introduced to 150 mL of 0.500 molarity potassium chromate, how many grams of silver chromate will precipitate? (12 points)

____ AgNO3(aq) + ____K2CrO4(aq) → ____Ag2CrO4(s)+ ____KNO3(aq)

Answer 1

2 AgNO_3(aq) + K₂CrO_4(aq) → Ag₂CrO_4(s) + 2KNO_3(aq)

C_M = n / V

n = C_M x V

n (AgNO3) = C_M x V = 0.700 x 0.2 = 0.14 mol

n (K₂CrO4) = C_M x V = 0.500 x 0.15 = 0.075 mol

According to the equation, 2 n (AgNO_3(aq)) = n (K₂CrO4) = n (Ag₂CrO4). Therefore, K₂CrO4 is the limiting reactant.

n (Ag₂CrO4) = n (K₂CrO4) = 0.075 mol

n = m / M

M (Ag₂CrO4) = 331.7 g/mol

m (Ag₂CrO4) = n x M = 0.075 x 331.7 = 24.9 g