Mg + 2HCl -> MgCl2 + H2 (1)
MgO + 2HCl -> MgCl2 + H2O (2)
nH2=0,05(mol)
Theo PTHH 1 ta có:
nMg=nH2=0,05(mol)
nHCl(1)=2nMg=0,1(mol)
mMg=24.0,05=1,2(g)
mMgO=9,2-1,2=8(g)
nMgO=\(\dfrac{8}{40}=0,2\left(mol\right)\)
Theo PTHH 2 ta có:
nHCl(2)=2nMgO=0,4(mol)
=>\(\sum\)nHCl=0,4+0,1=0,5(mol)
mHCl=36,5.0,5=18,25(g)
mdd HCl=18,25:14,6%=125(g)
b;
Theo PTHH 1 và 2 ta có:
nMgCl2=nMg=0,05(mol)
nMgCl2=nMgO=0,2(mol)
mMgCl2=95.0,25=23,75(g)
C% dd MgCl2=\(\dfrac{23,75}{9,2+125-0,05.2}.100\%=17,71\%\)
a) PTHH: Mg + 2HCl-----> MgCl2+H2(1)
MgO+2HCl-----> MgCl2+H2O(2)
Ta có : nH2=\(\dfrac{V}{22,4}\)=\(\dfrac{1,12}{22,4}\)=0,05(mol)
Theo PTHH(1): Ta có nMg=nH2=0,05(mol)
=> mMg=0,05 x 24=1,2 (g)
=> mMgO=9,2-1,2=8(g)
=> nMgO=\(\dfrac{8}{40}\)=0,2(mol)
Theo Pthh (1): nHCl=nMg=0,05(mol)
Pthh (2): nHCl=nMgO=0,2(mol)
=> nHCl= 0,05+0,2=0,25(mol)
=> mHCl= 0,25x36,5=9,125(g)
=> mddHCl=
