Question

Answer 1

\(1.\\ 2Al+6HCl->2AlCl_3+3H_2\\ Al_2O_3+6HCl->2AlCl_3+3H_2O\\ n_{Al}=\dfrac{2}{3}\cdot\dfrac{6,72}{22,4}=0,2mol\\ a.\%m_{Al}=\dfrac{0,2\cdot27}{36}=15\%\\ \%m_{Al_2O_3}=85\%\\ b.n_{Al_2O_3}=0,85\cdot\dfrac{36}{102}=0,3mol\\ C_M=\dfrac{0,6+1,8}{0,2}=12M\\ c.C_{M\left(AlCl_3\right)}=\dfrac{0,2+0,6}{0,2}=4\left(M\right)\)

Answer 2

\(2.\\ Zn+H_2SO_4->ZnSO_4+H_2\\ a.n_{Cu}=\dfrac{4,48}{22,4}=0,2mol\\ \%m_{Zn}=\dfrac{0,2\cdot65}{32,2}=40,37\%\\ \%m_{Cu}=59,63\%\\ b.C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,2}=1M\\ c.C_{M\left(ZnSO_4\right)}=\dfrac{0,2}{0,2}=1M\)