Срочно!!! Дам 80 балів

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Ответ:

дано

m(ppa BaCL2) = 160 g

w(BaCL2) = 10%

--------------------------

m(BaSO4)-?

m(BaCl2) = 160*10% / 100% = 16g

BaCL2+H2SO4-->BaSO4+2HCL

M(BaCL2) = 208 g/mol

n(BaCL2) = m/M = 16/ 208= 0.077 mol

n(BaCL2) = n(BaSO4) = 0.077 mol

M(BaSO4)= 233 g/mol

m(BaSO4) = n*M = 0.077* 233 = 17.941 g

ответ 17.941 г