\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\
pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,3
\(m_{Al}=0,2.27=5,4g\\
m_{HCl}=0,3.98=29,4g\\
m_{\text{dd}}=\dfrac{29,4.100}{30}=98g\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ pthh2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,3
\(m_{Al}=0,2.27=5,4g\\
m_{H_2SO_4}=\left(0,3.98\right).30\%=8,82g\)