Đáp án:
\(a) \; 1,12\text{(l)}\\ b)\;6\text{(g)}\\ c)\; 2,67\text{(g)}\)
Giải thích các bước giải:
\(a)\; n_{Mg}=\dfrac{1,2}{24}=0,05\text{(mol)}\\ n_{H_2SO_4}=\dfrac{7,35}{98}=0,075\text{(mol)}\\ PTHH:\; Mg+H_2SO_4\to MgSO_4+H_2\)
Vì $\dfrac{0,05}1<\dfrac{0,075}1\to$Sau phản ứng $Mg$ hết, $H_2SO_4$ dư
$n_{H_2}=n_{Mg}=0,05\text{(mol)}\to V_{H_2}=22,4\times 0,05=1,12\text{(l)}$
$b) \; n_{MgSO_4}=n_{Mg}=0,05\text{(mol)}\to m_{MgSO_4}=0,05\times 120=6\text{(g)}\\ c)\; n_{H_2}=n_{Mg}=0,05\text{(mol)}\\ 3H_2+Fe_2O_3\xrightarrow{t^{\circ}} 2Fe+ 3H_2O\\n_{Fe_2O_3}=\dfrac{0,05}3=\dfrac 1{60}\text{(mol)}\to m_{Fe_2O_3}=160\times \dfrac 1{60}=2,67\text{(g)}$