Bạn tham khảo:
$\text{a) PTHH: }$ $2Al + 3H_2SO_4→Al_2(SO_4)_3+3H_2$
$n_{Al}=\frac{5,4}{27}=0,2mol$
$n_{H_2SO_4}=\frac32.n_{Al}=\frac32.0,2=0,3mol$
$m_{H_2SO_4}=0,3.98=29,4g$
$⇒m_{ddH_2SO_4}=\frac{29,4}{9,8}.100=300g$
$b)$ $n_{Al_2(SO_4)_3}=\frac{1}{2}.n_{Al}=\frac12.0,2=0,1mol$
$⇒m_{Al_2(SO_4)_3}=0,1.342=34,2g$
$n_{H_2}=\frac32.n_{Al}=\frac32.0,2=0,3mol$
$⇒m_{H_2}=0,3.2=0,6g$
$m_{ddAl_2(SO_4)_3}=300+5,4-0,6=304,8g$
$\text{C%}_{Al_2(SO_4)_3}=\frac{34,2}{304,8}.100=11,22\text%$