a,
$n_{BaSO_4}=\dfrac{93,2}{233}=0,4(mol)$
$X(Al, M)+H_2SO_4\to$ muối sùnat $+H_2$
Muối sunfat $+Ba(NO_3)_2\to$ muối nitrat $+BaSO_4$
Bảo toàn $S$:
$n_{H_2SO_4}=n_{BaSO_4}=0,4(mol)$
Bảo toàn $H$:
$n_{H_2}=n_{H_2SO_4}=0,4(mol)$
$\to V=0,4.22,4=8,96l$
Bảo toàn $Ba$:
$n_{Ba(NO_3)_2}=n_{BaSO_4}=0,4(mol)$
BTKL:
$m_{\text{muối sunfat}}=57,4+93,2-0,4.261=46,2g$
BTKL:
$m=46,2+0,4.2-0,4.98=7,8g$
b,
Đặt $x$, $y$ là mol $Al$, $M$
Có $3n_{Al}+2n_M=2n_{H_2}$
$\to 1,5x+y=0,4$
$\to 27x+18y=7,2$
Mà $27x+M_M.y=7,8$
$\to y(M_M-18)=0,6$
$\to y=\dfrac{0,6}{M_M-18}(mol)$
Mà $1,5x+y=0,4\to x=\dfrac{0,4-y}{1,5}>0$
$\to y<0,4$
$\to M_M>19,5$
$y>\dfrac{1}{3}x\to x<3y$
$\to \dfrac{0,4-y}{1,5}<3y$
$\to 4,5y>0,4-y$
$\to y>\dfrac{4}{55}$
$\to M_M<26,25$
Do đó $19,5<M_M<26,25$
$\to M_M=24(Mg)$