When mixed, solutions of barium chloride, BaCl2, and potassium chromate, K2CrO4, form a yellow precipitate of barium chromate, BaCrO4. The balanced equation is: BaCl2 (aq) K2CrO4 (aq) → BaCrO4 (s) 2KCl (aq) How many grams of barium chromate are expected when a solution containing 0.41 mol BaCl2 is mixed with a solution containing 0.24 mol K2CrO4?

60.80 g

Explanation:

BaCl₂ (aq) + K₂CrO₄ (aq) → BaCrO₄ (s) + 2 KCl (aq)

For the reaction, given in the question,

1 mol of BaCl₂ reacts with 1 mol of K₂CrO₄ to give the corresponding products.

Hence, from the question,

0.41 mol of BaCl₂ requires 0.41 mol of K₂CrO₄

But, only 0.24mol of K₂CrO₄ are available,

Therefore,

K₂CrO₄ act as a limiting reagent, and will decide the product of the reaction,

Hence, from the reaction,

1 mol K₂CrO₄ gives, 1 mol BaCrO₄,

Hence,

0.24 mol of K₂CrO₄, will give 0.24 mol of BaCrO₄.

Moles is denoted by given mass divided by the molecular mass,

Hence,

n = w / m

n = moles,

w = given mass,

m = molecular mass.

From the question,

0.24 mol of BaCrO₄,

therefore,

m = molecular mass of BaCrO₄ = 253.37 g/mol

n = w / m

w = n * m

w = 0.24 mol * 253.37 g/mol = 60.80 g