Question

Дано m(BaCl2)=14г найти m(BaSo4)-? срочно​

Answer 1

Ответ:

ВаСl₂ + Na₂SO₄ = 2NaCl + BaSO₄↓

n(ВаСl₂) = m(ВаСl₂) /M(ВаСl₂) = 14г/ 208г/моль = 0,067моль

n(BaSO₄) = n(ВаСl₂) = 0,067моль

m(BaSO₄) = n(BaSO₄) * M(BaSO₄)

m(BaSO₄) = 0.067моль * 233г/моль = 15,6г

Объяснение:

Answer 2

Calculation of m(BaSo4) given m(BaCl2) = 14g

To calculate the mass of BaSO4 (barium sulfate) given the mass of BaCl2 (barium chloride), we need to use the balanced chemical equation for the reaction between BaCl2 and Na2SO4 (sodium sulfate):

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

From the equation, we can see that one mole of BaCl2 reacts with one mole of BaSO4. The molar mass of BaCl2 is 208.23 g/mol, and the molar mass of BaSO4 is 233.38 g/mol.

To find the mass of BaSO4, we can use the following equation:

m(BaSo4) = (m(BaCl2) * Molar mass of BaSO4) / Molar mass of BaCl2

Substituting the given values:

m(BaSo4) = (14g * 233.38 g/mol) / 208.23 g/mol

Calculating the result:

m(BaSo4) = 15.72g

Therefore, the mass of BaSO4 is 15.72 grams when the mass of BaCl2 is 14 grams.