PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (1)

             \(MgO+2HCl\rightarrow MgCl_2+H_2O\)  (2)

a) Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=n_{Mg}\) \(\Rightarrow m_{Mg}=1\cdot24=24\left(g\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{24}{32}\cdot100\%=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)

b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\dfrac{32-24}{40}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=2,4mol\) \(\Rightarrow m_{ddHCl}=\dfrac{2,4\cdot36,5}{7,3\%}=1200\left(g\right)\)

c) Theo PTHH: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=1,2mol\)

\(\Rightarrow\Sigma m_{MgCl_2}=1,2\cdot95=114\left(g\right)\)

Mặt khác: \(m_{H_2}=1\cdot2=2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ m_{Cu}=9,7-6,5=3,2g\\ b)C_{\%ZnCl_2}=\dfrac{0,1.136}{6,5+120-0,1.2}\cdot100=10,77\%\)

a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

                \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

b) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Zn}\) \(\Rightarrow n_{ZnO}=\dfrac{20-0,1\cdot65}{81}=\dfrac{1}{6}\left(mol\right)\)

\(\Rightarrow n_{ZnCl_2}=n_{Zn}+n_{ZnO}=\dfrac{4}{15}\left(mol\right)\) 

Mặt khác: \(m_{H_2}=0,1\cdot2=0,2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=119,8\left(g\right)\) \(\Rightarrow C\%_{ZnCl_2}=\dfrac{\dfrac{4}{15}\cdot136}{119,8}\cdot100\%\approx30,27\%\)

c) Giả sử khí là SO₂

PTHH: \(Zn+H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}ZnSO_4+SO_2\uparrow+H_2O\)

Theo PTHH: \(n_{SO_2}=n_{Zn}=0,1\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\)

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

           0,15       0,3          0,15

a) \(n_{Mg}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Mg}=0,1.24=2,4\left(g\right)\)

\(m_{MgO}=8,4-2,4=6\left(g\right)\)

⁰/₀Mg = \(\dfrac{2,4.100}{8,4}=28,57\)⁰/₀

⁰/₀MgO = \(\dfrac{6.100}{8,4}=71,43\)⁰/₀

b) Có : \(m_{MgO}=6\left(g\right)\)

\(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,2+0,3=0,5\left(mol\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)

\(n_{MgCl2\left(tổng\right)}=0,1+0,15=0,25\left(mol\right)\)

⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)

\(m_{ddspu}=8,4+500-\left(0,1.2\right)=508,2\left(g\right)\)

\(C_{MgCl2}=\dfrac{14,25.100}{508,2}=2,8\)⁰/₀

 Chúc bạn học tốt

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,2(mol)\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow m_{Cu}=20-5,6=14,4(g)\\ c,\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\ \%m_{Cu}=100\%-28\%=72\%\)