VD3:
$n_{Cl_2}=\dfrac{6,72}{22,4}=0,3 mol$
$4R+3O_2\buildrel{{t^o}}\over\to 2R_2O_3$
$\Rightarrow n_{R_2O_3}=0,2 mol$
$M_{R_2O_3}=\dfrac{20,4}{0,2}=102=2R+16.3$
$\Leftrightarrow R=27(Al)$
VD4:
$n_{H_2SO_4}=0,5.0,28=0,14 mol$
$n_{HCl}=0,5 mol$
$n_{H_2}=\dfrac{8,736}{22,4}=0,39 mol$
Ta thấy $0,5n_{HCl}+n_{H_2SO_4}=n_{H_2}$ nên phản ứng vừa đủ
BTKL: $m=7,74+0,14.98+0,5.36,5-0,39.2=38,93g$