Question

M(BaCl2)=100 г +H2SO4 m(осаду)-? BaCl2+H2SO4->BaSO4+2HCL

Answer 1

Дано:
m (BaCl2) = 100г

Найти:
m (BaSO4) - ?

Решение:
BaCl2 + H2SO4 = BaSO4 + 2HCl

n (BaCl2) = m/M = 100 / 208 = 0.48 моль

n (BaSO4) = n (BaCl2) = 0.48 моль (по уравнению)

m (BaSO4) = n * M = 0.48 * 233 = 111.84г

Ответ: m (BaSO4) = 111.84г

Answer 2

M(BaCl2) - молярная масса данного вещества =  208г/моль
n( BaCl2) = m/M= 100/208 ≈ 0,5 моль = n(BaSO4)
M(BaSO4) = 233 г/моль
m(BaSO4) = M*n = 233* 0,5 = 116,5 г

Answer 3

To calculate the mass of the precipitate formed in the reaction between barium chloride (BaCl2) and sulfuric acid (H2SO4), you need to identify the limiting reactant first. The limiting reactant is the one that is completely consumed, and it determines the maximum amount of product that can be formed.

The balanced chemical equation for the reaction is: BaCl2 + H2SO4 -> BaSO4 + 2HCl

Given:

• Mass of BaCl2 = 100 g

We need more information to proceed with the calculation. Specifically, we need to know the amount or mass of H2SO4 used in the reaction. Without this information, it is not possible to determine the limiting reactant and, consequently, the mass of the precipitate formed.

If you have the mass of H2SO4 or any other relevant information, please provide it, and I'd be happy to help you with the calculation.