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![](https://rs.olm.vn/images/avt/0.png?1311)
Dạ đề e viết sai khúc cuối 😅
Đốt cháy hoàn toàn hỗn hợp CH4 và C3H8 trong khí oxi,sau pứ thu dc 12,6 gam nước tính % V và %m hỗn hợp đồng
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi số mol Al, Na trong a gam hỗn hợp là x, y (mol)
=> 27x + 23y = a (1)
PTHH: 4Al + 3O2 --to--> 2Al2O3
x---------------->0,5x
4Na + O2 --to--> 2Na2O
y---------------->0,5y
=> 102.0,5x + 62.0,5y = 1,64.a
=> 51x + 31y = 1,64a (2)
(1)(2) => 51x + 31y = 1,64(27x + 23y)
=> 6,72x = 6,72y
=> x = y
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27x}{27x+23y}.100\%=54\%\\\%m_{Na}=\dfrac{23y}{27x+23y}.100\%=46\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,Đặt:n_{CH_4}=a\left(mol\right);n_{C_4H_{10}}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ 2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\\ \Rightarrow\left\{{}\begin{matrix}16a+58b=7,4\\22,4a+22,4.4b=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{CH_4}=0,1.16=1,6\left(g\right)\\m_{C_4H_{10}}=0,1.58=5,8\left(g\right)\end{matrix}\right.\\ b,n_{O_2}=2a+\dfrac{13}{2}b=2.0,1+6,5.0,1=0,85\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,85.22,4=19,04\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
Theo đề: \(m_{hh}=39\left(g\right)\)
\(\Rightarrow m_{Al}+m_{Fe}=39\\ \Rightarrow27x+56y=39\left(1\right)\)
\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ \left(mol\right)....x\rightarrow..0.75x....0,5x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow..\dfrac{2}{3}y.....\dfrac{1}{3}y\)
Theo đề: \(n_{O_2}=\dfrac{V}{22,4}=\dfrac{12,32}{22,4}=0,55\left(mol\right)\)
\(\Rightarrow0,75x+\dfrac{2}{3}y=0,55\left(2\right)\)
\(\xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}27x+56y=39\\0,75x+\dfrac{2}{3}y=0,55\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,6.56=33,6\left(g\right)\end{matrix}\right.\\ m_r=m_{Al_2O_3}+m_{Fe_3O_4}=0,5.0,2.102+\dfrac{1}{3}.0,6.232=56,6\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo ĐLBT KL, có: mKL + mO2 = m oxit
⇒ mO2 = 28,4 - 15,6 = 12,8 (g)
\(\Rightarrow n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al} = a ; n_{Fe} = b\Rightarrow 27a + 56b = 27,6(1)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,5a(mol)\\ n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = \dfrac{b}{3}(mol)\\ \Rightarrow 0,5a.102 + \dfrac{b}{3}232 = 43,6(2)\\ (1)(2) \Rightarrow a = 0,4 ; b = 0,3\\ m_{Al} = 0,4.27 = 10,8(gam) ; m_{Fe} = 0,3.56 = 16,8(gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phản ứng xảy ra:
\(2Mg+O_2\rightarrow^{t^o}=2MgO\)
\(4Al+3O_2\rightarrow^{t^o}2Al_2O_3\)
Gọi số mol của Mg là x; số mol của Al là y
\(\rightarrow m_{hh}=m_{Mg}+m_{Al}=24x+2yy=17,4g\)
Có:
\(n_{MgO}=n_{Mg}=x\)
\(n_{Al_2O_3}=\frac{1}{2}n_{Al}=0,5y\)
\(\rightarrow m_Y=40x+102.0,5y=30,2g\)
\(\rightarrow\hept{\begin{cases}x=0,5\\y=0,2\end{cases}}\)
\(\rightarrow n_{MgO}=n_{Mg}=0,5mol\)
\(\rightarrow m_{MgO}=0,5.40=20g\)
Câu 1:
Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}n_{CuO}=n_{Cu}=b\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}a\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow102\cdot\dfrac{1}{2}a+80b=21,1\) (1)
Ta có: \(n_{O_2}=\dfrac{3,92}{22,4}=0,175\left(mol\right)\)
Bảo toàn electron: \(3a+2b=0,7\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,1\cdot27=2,7\left(g\right)\\m_{Cu}=0,2\cdot64=12,8\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{2,7+12,8}\cdot100\%\approx17,42\%\\\%m_{Cu}=82,58\%\end{matrix}\right.\)