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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?

Respuesta :

Answer:

60.80 g

Explanation:

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

For the reaction , given in the question ,

1 mol of BaClâ‚‚ reacts with 1 mol of Kâ‚‚CrOâ‚„ 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 Kâ‚‚CrOâ‚„ are available ,

Therefore ,

Kâ‚‚CrOâ‚„ 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 Kâ‚‚CrOâ‚„ , will give 0.24 mol of BaCrOâ‚„ .

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 BaCrOâ‚„ ,

therefore ,

m = molecular mass of BaCrOâ‚„ = 253.37 g/mol

n = w / m

w = n * m

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

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