a) Mg + 2HCl --> MgCl2 + H2
MgO + 2HCl --> MgCl2 + H2O
b) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
______0,1<--0,2<------0,1<---0,1_____(mol)
=> mMg = 0,1.24 = 2,4 (g)
=> mMgO = 4,4-2,4 = 2 (g)
c) \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
PTHH: MgO + 2HCl --> MgCl2 + H2O
______0,05-->0,1------->0,05
=> nHCl = 0,1 + 0,2 = 0,3 (mol)
=> mHCl = 0,3.36,5 = 10,95(g)
=> \(m_{dd}=\dfrac{10,95.100}{7,3}=150\left(g\right)\)
d) mMgCl2 = 95(0,1+0,05) = 14,25 (g)
mdd = 4,4 + 150 - 0,1.2 = 154,2(g)
=> \(C\%\left(MgCl_2\right)=\dfrac{14,25}{154,2}.100\%=9,24\%\)
Câu 5 :
\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,1 0,2 0,1 0,1
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,05 0,1 0,05
b) \(n_{Mg}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{Mg}=0,1.24=2,4\left(g\right)\)
\(m_{MgO}=4,4-2,4=2\left(g\right)\)
c) Có : \(m_{MgO}=2\left(g\right)\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,2+0,1=0,3\left(mol\right)\)
\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
\(m_{ddHCl}=\dfrac{10,95.100}{7,3}=150\left(g\right)\)
d) \(n_{MgCl2\left(tổng\right)}=0,1+0,05=0,15\left(mol\right)\)
⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)
\(m_{ddspu}=4,4+150-\left(0,1.2\right)=154,2\left(g\right)\)
\(C_{MgCl2}=\dfrac{14,25.100}{154,2}=9,24\)0/0
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