Đáp án:
Mg + 2HCl → $MgCl_{2}+H_{2}↑$
MgO + 2HCl→ $MgCl_{2}+H_{2}O$
a) Ta có: $n_{H_{2}}$=0,05 (mol)
⇒$n_{Mg}$=0,05 (mol)
⇒$m_{Mg}$=0,05.24=1,2 (gam)
⇒%Mg=$\frac{1,2}{9,2}.100$=13,04%
⇒%MgO=100-13,04=86,96%
b) $m_{MgO}$=9,1-1,2=8 (g)
$\Rightarrow n_{MgO}=0,2 (mol)$
$\Rightarrow n_{HCl}$=2.0,05+0,2.2=0,5 (mol)
$\Rightarrow m_{HCl}$=0,5.36,5=18,25 (gam)
⇒$m_{dd_{HCl}}=\frac{18,25.100}{19,26}$=94,76 (gam)
c) $n_{MgCl_{2}}$=0,05+0,2=0,25(mol)
$\Rightarrow m_{MgCl_{2}}$=0,25.95=23,75 (gam)
C%=$\frac{23,75.100}{94,76}$=25,06%