Mg + 2HCl -> MgCl2 + H2 (1)
MgO + 2HCl -> MgCl2 + H2O (2)
nH2=0,05(mol)
Từ 1:
nMg=nH2=0,05(mol)
mMg=24.0,05=1,2(g)
=>mMgO=9,2-1,2=8(g)\(\Leftrightarrow\)0,2(mol)
Theo PTHH 1 và 2 ta có:
nHCl(1)=2nMg=0,1(mol)
nHCl(2)=2nMgO=0,4(mol)
nMgCl2=\(\dfrac{1}{2}\)nHCl=0,25(mol)
mHCl=36,5.0,5=18,25(g)
mdd HCl=18,25:14,6%=125(g)
mdd=125+9,2-2.0,05==134,1(g)
mMgCl2=95.0,25=23,75(g)
C% dd MgCl2=\(\dfrac{23,75}{134,1}.100\%=17,71\%\)
Chỉ có Mg tác dụng với dung dịch HCl tạo ra khí H2
=> nH2=0,05(mol)
PT: Mg + 2HCl -> MgCl2 + H2
vậy: 0,05<--0,1<-----0,05<---0,05(mol)
=> mMg=n.M=0,05.24=1,2(g)
=> mMgO=mhh-mMg=9,2-1,2=8(g)
=> nMgO=0,2(mol)
Pt2: MgO + 2HCl -> MgCl2 + H2O
vậy: 0,2------>0,4------>0,2 (mol)
Ta có: \(m_{HCl}=\left(0,4+0,1\right).36,5=18,25\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
mMgCl2=(0,2+0,05).95=23,75(g)
\(m_{ddsauphanung}=m_{hh}+m_{ddHCl}-m_{H_2}=9,2+100-\left(0,05.2\right)=109,1\left(g\right)\)
\(\Rightarrow C\%_{ddsauphanung}=\dfrac{m_{MgCl_2}.100\%}{m_{ddsauphanung}}=\dfrac{23,75.100}{109,1}\approx21,77\left(\%\right)\)
a. Pt1 - Mg + 2HCl --> MgCl2 + H2
0.05 0.1 0.05 0.05
Pt2 - MgO + 2HCl ---> MgCl2 + H2O
0.2 0.4 0.2
nH2 = \(\dfrac{1.12}{22.4}=0.05mol\)
b. Suy ra nMg =0.05 mol.
---> mMg = 0.05 * 24 = 1.2 g.
---> mMgO = 9.2 - 1.2 = 8g.
c. ---> nMgO = \(\dfrac{8}{40}=0.2mol\)
---> nMgCl2 = 0.2 mol.
=> nMgCl2 ở 2 pt = 0.05 + 0.2 = 0.25 mol.
=> mMgCl2 = 23.75 g.
nHCl = 0.1 + 0.4 = 0.5 mol.
=> mHCl = 0.5*36.5 = 18.25 g.
=> m dd HCl = \(\dfrac{18.25\cdot100}{14.6}=125g\)
mH2 = 0.05*2 = 0.1 g
=> m dd spứ = 9.2 + 125 - 0.1 = 134.1 g
Nồng độ phần trăm dd MgCl2 là
\(\dfrac{18.25\cdot100}{134.1}=13.6\text{%}\)
