PTHH: Mg + 2HCl \(\rightarrow\) MgCl2 + H2\(\uparrow\) (1)
Theo pt: 1......... 2 ........... 1 ........ 1 ... (mol)
Theo đề: 0,1 ... 0,2 ........ 0,1 ..... 0,1 .. (mol)
PTHH: MgO + 2HCl \(\rightarrow\) MgCl2 + H2O (2)
Theo pt: . 1......... 2 ............ 1 ......... 1 ... (mol)
Theo đề: 0,09 ... 0,18 ...... 0,09 .... 0,09 . (mol)
1/ \(n_{H_2}=\dfrac{V_{đktc}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(m_{Mg}=n.M=0,1.24=2,4\left(g\right)\)
Ta có: mMg + mMgO = 6 (gt)
\(\Rightarrow\) mMgO = 6 - mMg = 6 - 2,4 = 3,6 (g)
\(\%m_{Mg}=\dfrac{m_{Mg}}{m_{hh}}.100\%=\dfrac{2,4}{6}.100\%=40\%\)
\(\%m_{MgO}=\dfrac{m_{MgO}}{m_{hh}}.100\%=\dfrac{3,6}{6}.100\%=60\%\)
2/ \(n_{MgO}=\dfrac{m}{M}=\dfrac{3,6}{40}=0,09\left(mol\right)\)
\(n_{HCl}=n_{HCl\left(1\right)}+n_{HCl\left(2\right)}=0,2+0,18=0,38\left(mol\right)\)
\(m_{HCl}=n.M=0,38.36,5=13,87\left(g\right)\)
\(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{13,87.100\%}{20\%}=69,35\left(g\right)\)
\(V_{ddHCl}=\dfrac{m_{ddHCl}}{D}=\dfrac{69,35}{1,12}\approx61,92\left(ml\right)\approx0,062\left(l\right)\)
3/ \(m_{MgCl_2}=m_{MgCl_{2\left(1\right)}}+m_{MgCl_{2\left(2\right)}}=95\left(0,1+0,09\right)=18,05\left(g\right)\)
\(m_{H_2}=n.M=0,1.2=0,2\left(g\right)\)
\(m_{ddspứ}=m_{Mg}+m_{MgO}+m_{ddHCl}-m_{H_2}=6+69,35-0,2=75,15\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{m_{MgCl_2}}{m_{ddspứ}}.100\%=\dfrac{18,05}{75,15}.100\%\approx24,02\%\)
