Question

Answer 1

nH2 = 6,72/22,4 = 0,3 (mol)

PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2

nAl = 0,3 : 3 . 2 = 0,2 (mol)

mAl = 0,2 . 27 = 5,4 (g)

Answer 2

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)

\(\rightarrow n_{Al}=0,2mol\Rightarrow m_{Al}=n.M=0,2.27=5,4g\)