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EstequiometriaFGVQuímica Geral

Qual a massa necessária de Ca(OH)2 para reagir…

(FGV-­SP) Balanceando­-se a reação

Aℓ2(SO4)3 + 3 Ca(OH)2 → y Aℓ(OH)3 + z CaSO4

corretamente, obtêm­-se os valores de y e z. Qual a massa necessária de Ca(OH)2 para reagir completamente com w quilogramas de Aℓ2(SO4)3?

Dados:

Massas molares de Aℓ2(SO4)3 = 342 g/mol, Ca(OH)2 = 74 g/mol e w = y + z

a) 3,25 quilogramas.

b) 5,40 quilogramas.

c) 4,62 quilogramas.

d) 1,08 quilogramas.

e) 67,57 gramas.

Utilize a tabela periódica para calcular as massas molares quando necessário.

Resolução

Equação química balanceada:

Aℓ2(SO4)3 + 3 Ca(OH)2 → 2 Aℓ(OH)3 + 3 CaSO4

Logo, y = 2, z = 3 e w = 5.

De acordo com a equação balanceada, 1 mol de Aℓ2(SO4)3 reage com 3 mols de Ca(OH)2. Assim, temos que a massa de Ca(OH)2 necessária para reagir com 5 kg de Aℓ2(SO4)3 é dada por:

Gabarito: LETRA A

Confira mais EXERCÍCIOS RESOLVIDOS sobre ESTEQUIOMETRIA.

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