\(a.n_{Ba}=0,4\left(mol\right);n_{CuSO_4}=0,15\left(mol\right)\\ Ba+H_2O\rightarrow Ba\left(OH\right)_2+H_2\left(1\right)\\ Ba\left(OH\right)_2+CuSO_4\rightarrow BaSO_4+Cu\left(OH\right)_2\left(2\right)\\ KhíA:H_2\\ n_{H_2}=n_{Ba}=0,4\left(mol\right)\\ \Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b.LTLpứ\left(2\right):\dfrac{0,4}{1}>\dfrac{0,15}{1}\Rightarrow Ba\left(OH\right)_2dư\\ DungdịchC:Ba\left(OH\right)_2\\ C\%_{Ba\left(OH\right)_2}=\dfrac{\left(0,4-0,15\right).171}{54,8+400-0,15.233-0,15.98-0,4.2}.100=10,57\%\\ c.KếttủaB:BaSO_4,Cu\left(OH\right)_2\\ NungB:Cu\left(OH\right)_2-^{t^o}\rightarrow CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=0,15\left(mol\right)\\ \Rightarrow m_{CuO}=0,15.80=12\left(g\right)\\ m_{BaSO_4}=0,15.233=34,95\left(g\right)\\ H=\dfrac{12+34,95}{47,49}=98,86\%\)
