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Ответ:

дано

m(Al2(SO4)3) =3.42 g

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m(Ba(OH)2)-?

3Ba(OH)2+Al2(SO4)3-->2Al(OH)3+3BaSO4

M(Al2(SO4)3)= 342 g/mol

n(Al2(SO4)3) = m/M = 3.42 / 342 = 0.01 mol

3n(Ba(OH)2) = n(Al2(SO4)3)

n(Ba(OH)2 = 3*0.01 / 1 = 0.03 mol

M(Ba(OH)2) = 171 g/mol

m(Ba(OH)2) = n*M =0.03 * 171 = 5.13 g

ответ 5.13 г

Объяснение: