Đáp án:

Giải thích các bước giải:

a. PTHH : $2Al + 3H_2SO4 → Al_2(SO_4)_3 + 3H_2$
b.

có : $n_{Al} = \frac{5,4}{27} = 0,2(mol)$

Theo PT , $n_{Al2(SO4)3} = \frac{1}{2}n_{Al} = 0,1(mol)$

$⇒ m_{muối} = 0,1.342 = 34,2(gam)$

Theo PT , $n_{H2} = n_{Al}.\frac{3}{2} = 0,3(mol)$
$⇒ V_{H2} = 0,3.22,4 = 6,72(lít)$

c.

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$

Theo PT , $n_{Fe} = n_{H2}.\frac{2}{3} = 0,3.\frac{2}{3} = 0,2(mol)$
$⇒ m_{Fe} = 0,2.56 = 11,2(gam)$

$a,PTPƯ:2Al+3H_2SO_4\xrightarrow{} Al_2(SO_4)_3+3H_2↑$

$Al_2(SO_4)_3:$ $\text{Nhôm sunfat}$

$H_2:$ $\text{Hiđrô}$

$b,n_{Al}=\dfrac{5,4}{27}=0,2mol.$

$Theo$ $pt:$ $n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1mol.$

$⇒m_{Al_2(SO_4)_3}=0,1.342=34,2g.$

$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol.$

$⇒V_{H_2}=0,3.22,4=6,72l.$

$c,PTPƯ:Fe_2O_3+3H_2\xrightarrow{t^o} 2Fe+3H_2O$

$Theo$ $pt:$ $n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2mol.$

$⇒m_{Fe}=0,2.56=11,2g.$

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