Ba + 2H2O-----.>Ba(OH)2 +H2(1)
Ba(OH)2 + FeSO4-----> BaSO4 + Fe(OH)2(1)
4Fe(OH)2 +O2-----> 2Fe2O3+4H2O(3)
Ta có
n\(_{Ba}=\frac{27,4}{137}=0,2\left(mol\right)\)
Theo pthh1
n\(_{Ba\left(OH\right)2}=n_{Ba}=0,2\left(mol\right)\)
Theo pthh2
n\(_{Fe\left(OH\right)2}=n_{Ba\left(OH\right)2}=0,2\left(mol\right)\)
Theo pthh3
n\(_{Fe2O3}=\frac{1}{2}n_{Fe\left(OH\right)2}=0,1\left(mol\right)\)
x=m\(_{Fe2O3}=0,1.160=16\left(g\right)\)
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