Answer:
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
