Verified answer

дано

m(Ba(OH)2) = 342 g

m(HCL) = 73 g

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m(BaCL2)-?

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

n(Ba(OH)2)  = m/M = 342 / 171 = 2 mol

M(HCL) = 36.5 g/mol

n(HCL) = m/M = 73 / 36.5 = 2 mol

n(Ba(OH)2) =0.5 n(HCL)

соляная кислота в недостатке

0.5n(HCL) = n(BaCL2)

Ba(OH)2+2HCL --> BaCL2↓+2H2O

M(BaCL2) = 208 g/mol

m(BaCL2) = n*M = 0.5*208 = 104 g

ответ 104 г