How many grams of sodium chromate, Na2CrO4, are needed to react completely with
56.7g of silver nitrate, AgNO3?

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

Explanation:

Step 1: data given

Mass of silver nitrate AgNO3 = 56.7 grams

Molar mass AgNO3 = 169.87 g/mol

Molar mass of Na2CrO4 = 161.97 g/mol

Step 2: The balanced equation

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 56.7 grams / 169.87 g/mol

Moles AgNO3 = 0.334 moles

Step 4: Calculate moles Na2CrO4 moles needed

For 2 moles AgNO3 we need 1 mol Na2CrO4 to produce 1 mol Ag2CrO4 and 2 moles NaNO3

For 0.334 moles AgNO3 we need 0.334 / 2 = 0.167 moles Na2CrO4

Step 5: Calculate mass Na2CrO4

Mass Na2CrO4 = 0.167 moles * 161.97 g/mol

Mass Na2CrO4 = 27.0 grams

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

dynamic equilibrium is the answer.