nAl=\(\dfrac{5,4}{27}=0,2mol\)
\(pt:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(n_{p\text{ư}}:0,2\rightarrow0,3\) \(\rightarrow0,1\) \(\rightarrow0,3\)
a)VH2=0,3.22,4=6,72l
b)mH2SO4=0,3.98=29,4g
mddH2SO4=\(\dfrac{29,4.100\%}{10\%}=294g\)
c)mdd Al2(SO4)3=mAl+mddH2SO4
=5,4+294=299,4g
mAl2(SO4)3=0,1.342=34,2g
C%\(\left[Al_2\left(SO_4\right)_3\right]=\dfrac{34,2}{299,4}.100\%\simeq11,423\%\)