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