Ответ:

дано

m(ppa H2SO4) = 400 g

W(H2SO4) = 12%

m(BaCL2) = 300 g

W(BaCL2) = 15%

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m(BaSO4) - ?

m(H2SO4) = 400 *12% / 100% = 48 g

M(H2SO4) = 98 g/mol

n(H2SO4) = m/M = 48 / 98 = 0.5 mol

m(BaCL2) = 300*15% / 100% = 45 g

M(BaCL2) = 208 g/mol

n(BaCL2) = m/M = 45 / 208 = 0,22 mol

n(H2SO4) > n(BaCL2)

H2SO4+BaCL2-->2HCL+BaSO4

n(BaCL2) = n(BaSO4) = 0.22 mol

M(BaSO4) = 233 g/mol

m(BaSO4) = n*M = 0.22 * 233 = 51.26 g

ответ 51.26 г

Объяснение: