a) Fe+H2SO4----->FeSO4 +H2
Ta có
m\(_{H2SO4}=\frac{200.14,6}{100}=29,2\left(g\right)\)
n\(_{H2SO4}=\frac{29,2}{98}=0,3\left(mol\right)\)
Theo pthh
n\(_{Fe}=n_{H2SO4}=0,3\left(mol\right)\)
a=m\(_{Fe}=0,3.56=16,8\left(mol\right)\)
b) Theo pthh
n\(_{H2}=n_{H2SO4}=0,3\left(mol\right)\)
V\(_{H2}=0,3.22,4=6,72\left(l\right)\)
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