5/
$nH_2=\frac{8,96}{22,4}=0,4$
Ta có :
$mMg+mAl=7,8$
⇒$24nMg+27nAl=7,8(1)$
$Mg^{0} \to Mg^{2+}+2e$
$Al^{0} \to Al^{3+}+3e$
$H_2^{+1}+2e \to H_2^{0}$
Bảo toàn e ta có $ne_{cho}=ne_{nhận}$
⇒$2nMg+3nAl=2nH_2=0,8(2)$
(1)(2)⇒$\left \{ {{nMg=0,1} \atop {nAl=0,2}} \right.$
$\%mMg=\frac{0,1.24}{7,8}.100=30,77\%$
$\%mAl=100-30,77=69,23\%$
Bảo toàn nguyên tố "H" ⇒$nH_2SO_4=nH_2=0,4$
$VddH_2SO_4=\frac{0,4}{1,2}=\frac{1}{3}lit$
6/
$nH_2=\frac{3,36}{22,4}=0,15$
Ta có :
$mMg+mFe=6,8$
⇒$24nMg+56nFe=6,8(1)$
$Mg^{0} \to Mg^{2+}+2e$
$Fe^{0} \to Fe^{2+}+2e$
$H_2^{+1}+2e \to H_2^{0}$
Bảo toàn e ta có $ne_{cho}=ne_{nhận}$
⇒$2nMg+2nFe=2nH_2=0,3(2)$
(1)(2)⇒$\left \{ {{nMg=0,05} \atop {nFe=0,1}} \right.$
$\%mMg=\frac{0,05.24}{6,8}.100=17,65\%$
$\%mFe=100-17,65=82,35\%$
b,
$Mg^{0} \to Mg^{2+}+2e$
$Fe^{0} \to Fe^{2+}+2e$
$S^{+6}+2e \to S^{+4}$
Bảo toàn e ta có $ne_{cho}=ne_{nhận}$
⇒$2nMg+2nFe=2nSO_2$
⇒$nSO_2=0,15$
$VSO_2=0,15.22,4=3,36lit$
