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![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: x 1,5x
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)
b)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,1 0,15 0,05
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,1 0,1 0,1
\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)
mdd sau pứ = 5,1+250-0,15.2 = 254,8(g)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)
\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}27a+24b=5,1\\1,5a+b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,1}.100\approx52,941\%\\\%m_{Mg}\approx47,059\%\end{matrix}\right.\)
\(b.m_{ddH_2SO_4}=\dfrac{0,25.98.100}{9,8}=250\left(g\right)\\ m_{ddsau}=m_{Al,Mg}+m_{ddH_2SO_4}-m_{H_2}=5,1+250-0,25.2=254,6\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342}{254,6}.100\approx6,716\%\\ C\%_{ddMgSO_4}=\dfrac{0,1.120}{254,6}.100\approx4,713\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(m_{H_2}=1\cdot2=2\left(g\right)\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ Mg+H_2SO_4\to MgSO_4+H_2\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{Mg}=0,25(mol)\\ a,\begin{cases} \%_{Mg}=\dfrac{0,25.24}{14}.100\%=42,86\%\\ \%_{MgO}=100\%-42,86\%=57,14\% \end{cases}\\ b,n_{MgO}=\dfrac{14-0,25.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{H_2SO_4}=0,2+0,25=0,45(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{0,45.98}{200}.100\%=22,05\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)
b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)
\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)
Bạn tham khảo nhé!
a, nH2 = 0,03 ( mol )
=> nMg = nH2 = 0,03 ( mol )
=> mMg = 0,72 g
=> %Mg \(\approx\) 41,38 % .
=> % Al \(\approx\) 58,62 % .
b, Có : nH2 = 0,03 mol
=> nHCl = nHCltừ Al2O3 + nHCltừ Mg = 0,06 + 0,06 = 0,12 ( mol )
=> mHCl = 4,38 ( g )
Lại có : mdd = mhh + mddHCl = 501,74 ( g )
=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)
( chắc đoạn trên là Al2O3 :vvvv )
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