a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH :
$n_{H_2} = \dfrac{3}{2}n_{Al} = 0,3(mol) \Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)$
b) $n_{Al\ pư} = 0,2.90\% = 0,18(mol)$
$n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al\ pư} = 0,09(mol)$
$m_{Al_2(SO_4)_3} = 0,09.342 = 30,78(gam)$
c) $n_{H_2SO_4} = \dfrac{3}{2}n_{Al\ pư} = 0,27(mol)$
$m_{H_2SO_4} =0,27.98 = 26,46(gam)$
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