a,$MgO+2HCl→MgCl_2+H_2O$
$MgCO_3+2HCl→MgCl_2+H_2O+CO_2$
$nCO_2=4,48/22,4=0,2(mol)$
$⇒nMgCO_3=nCO_2=0,2(mol)$
$⇒mMgCO_3=0,2.84=16,8(g)$
$⇔mMgO=32,8-16,8=16(g)$
b,$nMgO=16/40=0,4(mol)$
$⇒nHCl=0,8+0,4=1,2(mol)$
$⇒mHCl=1,2.36,5=43,8(g)$
$⇒mddHCl=43,8.100/14,6=300g$
c,$nMgCl_2=0,4+0,2=0,6(mol)$
$⇒mMgCl_2=0,6.95=57(g)$
$mddspu=32,8+300-0,2.44=324(g)$
$⇒C$%$MgCl_2=57.100/324≈17,592$%