Bài giải:
a.
$n_{H_2}=\frac{2,24}{22,4}=0,1(mol)$
$Mg+2HCl→MgCl_2+H_2↑$
0,1 0,2 ← 0,1 (mol)
$MgO+2HCl→MgCl_2+H_2O$
0,09 → 0,18 (mol)
-$m_{Mg}=0,1.24=2,4(g)$
-$m_{MgO}=m_{hh}-m_{Mg}=6-2,4=3,6(g)$
$n_{MgO}=\frac{3,6}{40}=0,09(mol)$
-%$m_{Mg}=\frac{2,4}{6}.100$%$=40$%
-%$m_{MgO}=100$%$-$%$m_{Mg}=100$%$-40$%$=60$%
b.
$∑n_{HCl}=0,2+0,18=0,38(mol)$
$m_{chất..tan..HCl}=0,38.36,5=18,37(g)$
$⇒m_{dung..dịch..HCl}=\frac{18,37.100}{20}=69,35(g)$
Vì $d_{HCl}=1,1(g/ml)$
$⇒V_{HCl}=\frac{69,35}{1,1}=63,045(ml)$