Giải thích các bước giải:
a) `n_(Mg)=6/24=0,25(mol)`
`PT:Mg+H_2SO_4->MgSO_4+H_2`
Thep PT:
`n_(MgSO_4)=n_(H_2)=0,25(mol)`
`->V_(H_2)=0,25xx22,4=5,6(l)`
`m_(MgSO_4)=0,25xx120=30(g)`
b) `n_(Fe_2O_3)=32/160=0,2(mol)`
$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }PT\text{ }:\text{ }\text{ }Fe_2O_3+3H_2→2Fe+3H_2O\\\text{Trước p/ứ: }\text{ }\text{ }\text{ }\text{ }0,2\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0,25\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\text{ }(mol)\\\text{Khi p/ứ: }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{0,25}{3}\text{ }\text{ }\text{ }\text{ }\text{ }0,25\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{0,5}{3}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0,25\text{ }(mol)\\\text{Sau p/ứ: }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{7}{60}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac16\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0,25\text{ }(mol)$
`->Fe_2O_3` dư
Theo PT, `n_(Fe)=1/6(mol)`
`->m_(Fe)=1/6xx56=28/3(g)`
