Answer to Question #307650 in Chemistry for jack

Question

A chemist reacts 45.0 mL of 3.4 M AgNO₃ with an excess of Na₂CrO₄. How many grams of silver chromate will be produced?

Answer 1

Solution:

Molarity = Moles of solute / Liters of solution

Therefore,

Moles of AgNO₃ = Molarity of AgNO₃ × Liters of AgNO₃ solution

Moles of AgNO₃ = (3.4 M) × (45.0 mL) × (1 L / 1000 mL) = 0.153 mol

Balanced chemical equation:

2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃

According to stoichiometry:

2 mol of AgNO₃ produces 1 mol of Ag₂CrO₄

Thus, 0.153 mol of AgNO₃ produces:

Moles of Ag₂CrO₄ = (0.153 mol AgNO₃) × (1 mol Ag₂CrO₄ / 2 mol AgNO₃) = 0.0765 mol Ag₂CrO₄

The molar mass of Ag₂CrO₄ is 331.73 g/mol

Therefore,

Mass of Ag₂CrO₄ = (0.0765 mol Ag₂CrO₄) × (331.73 g Ag₂CrO₄ / 1 mol Ag₂CrO₄) = 25.377 g Ag₂CrO₄

Mass of Ag₂CrO₄ = 25.377 g Ag₂CrO₄ = 25.4 g Ag₂CrO₄

Answer: 25.4 grams of silver chromate (Ag₂CrO₄) will be produced