Question

How many grams of sodium chromate, Na2CrO4, are needed to react completely with

Answer 1

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