Bạn tham khảo:
$2R+3H_2SO_4 \to R_2(SO_4)_3+3H_2$
$n_{H_2}=\frac{13,44}{22,4}=0,6(mol)$
$n_R=\frac{2}{3}.0,6=0,4(mol)$
$a/ M_R=\frac{10,8}{0,4}=27(g/mol)$
$R=27(Al)$
$b/$
$n_{H_2SO_4}=n_{H_2}=0,6(mol)$
$m_{H_2SO_4}=0,6.98=58,8(g)$
$c/$
C1:
$2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2$
$n_{Al_2SO_4)_3}=0,2(mol)$
$m_{Al_2(SO_4)_3}=0,2.342=68,4(g)$
C2:
BTKL:
$m_{Al}+m_{H_2SO_4}=m_{Al_2(SO_4)_3}+n_{H_2}$
$10,8+58,8=m_{Al_2(SO_4)_3}+0,6.2$
$m_{Al_2(SO_4)_3}=68,4(g)$
