What mass of Na2CrO4 is required to precipitate all of the silver ions from 73.6 mL of a 0.150 M solution of AgNO3?

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