`n_{H_2}={7,437}/{24,79}=0,3(mol)`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`0,2←0,3←0,1←0,3(mol)`

`a=0,2.27=5,4(g)`

`b={0,3.98}/{20\%}=147(g)`

`c=0,1.342=34,2(g)`

$\rm 1)$

$\rm n_{H_2} = \dfrac{V}{24,79} = \dfrac{7,437}{24,79} = 0,3\ (mol)$

$\rm 2Al+3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2$

$\rm Theo\ pt:n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{2}{3} \times 0,3 = 0,2\ (mol)$

`->` $\rm a = m_{Al} = n \times M = 0,2 \times 27 = 5,4\ (g)$

$\rm Theo\ pt:n_{H_2SO_4} = n_{H_2} = 0,3\ (mol)$

$\rm m_{H_2SO_4} = n \times M = 0,3 \times 98 = 29,4\ (g)$

`->` $\rm b = m_{dd\ H_2SO_4} = \dfrac{m_{ct}}{C\%} \times 100 = \dfrac{29,4}{20} \times 100 = 147\ (g)$

$\rm Theo\ pt:n_{Al_2(SO_4)_3}= \dfrac{1}{3}n_{H_2} = \dfrac{1}{3} \times 0,3 = 0,1\ (mol)$

`->` $\rm c = m_{Al_2(SO_4)_3}= n \times M = 0,1 \times 342 = 34,2\ (g)$