Answer:
We need 0.894 grams of Na2CrO4
Explanation:
Step 1: Data given
Volume of a 0.150 M AgNO3 = 73.6 mL = 0.0736 L
Step 2: Calculate moles of Ag+
Moles Ag+ = moles AgNO3
Moles Ag+ = volume * molarity
moles Ag+ = 0.0736 L x 0.150 M = 0.01104 moles
Step 3: Calculate moles Na2CrO4
2AgNO3 + Na2CrO4 → Ag2CrO4 (s) + 2NaNO3
For 2 moles AgNO3 we need 1 mol Na2CRO4
For 0.01104 moles AgNO3 we need 0.01104/2 = 0.00552 moles Na2CrO4
Step 4: Calculate mass of Na2CrO4
Mass Na2CrO4 = moles * molar mass
Mass Na2CrO4 = 0.00552 moles * 161.97 g/mol
Mass Na2CrO4 = 0.894 grams
We need 0.894 grams of Na2CrO4
