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Sửa đề thành H2 nhé!
\(n_{SO_2}=x;n_{O_2}=y\\ V_{CH_4}=V_{O_2}\Rightarrow n_{CH_4}=n_{O_2}=y\)
Theo đề bài ta có:
\(d_{A/H}=\frac{M_A}{M_{H_2}}=16\Leftrightarrow M_A=16.2=32\)
\(\Rightarrow\sum_m=\overline{A}.\sum_n=32.\left(x+2y\right)\left(1\right)\)
\(\sum_V=22,4\left(x+2y\right)\left(l\right)\)
\(\sum_m=64x+32y+18y=64x+50y\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow32x+64y=64x+50y\\ \Leftrightarrow\frac{x}{y}=\frac{7}{16}\Rightarrow\left\{{}\begin{matrix}x=7\\y=16\end{matrix}\right.\)
\(\%V_{SO_2}=\frac{22,4x}{22,4\left(x+2y\right)}.100\%=\frac{x}{x+2y}.100\%=\frac{7}{7+16.2}.100\%=17,94\left(\%\right)\)
\(\%V_{O_2}=\%V_{CH_4}=\frac{22,4y}{22,4\left(x+2y\right)}.100\%=\frac{y}{x+2y}.100\%=\frac{16}{7+16.2}.100\%=41,03\left(\%\right)\)
\(\%m_{SO_2}=\frac{64x}{64x+50y}.100\%=\frac{64.7}{64.7+50.16}.100\%=35,9\left(\%\right)\)
\(\%m_{O_2}=\frac{32y}{64x+50y}.100\%=\frac{32.16}{64.7+50.16}.100\%=41,03\left(\%\right)\)
\(\%m_{SO_2}=\frac{18y}{64x+50y}.100\%=\frac{18.16}{64.7+50.16}.100\%=23,07\left(\%\right)\)
a. Gọi x, y lần lượt là số mol của CH4 và CO2
Ta có: \(n_A=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo đề, ta có:
- x + y = 0,4 (1)
- 16x + 44y = 9,2 (2)
Từ (1) và (2), ta có HPT:
\(\left\{{}\begin{matrix}x+y=0,4\\16x+44y=9,2\end{matrix}\right.\)
Giải ra, ta được:
x = 0,3, y = 0,1
=> \(m_{CH_4}=0,3.16=4,8\left(g\right);m_{CO_2}=0,1.44=4,4\left(g\right)\)
b. Ta có: \(\overline{M_A}=\dfrac{4,8+4,4}{0,3+0,1}=23\left(g\right)\)
=> \(d_{\dfrac{A}{O_2}}=\dfrac{\overline{M_A}}{M_{O_2}}=\dfrac{23}{32}=0,71875\left(lần\right)\)
- Xét A:
Giả sử \(m_{SO_2}=m_{CH_4}=16\left(g\right)\)
\(n_{SO_2}=\dfrac{16}{64}=0,25\left(mol\right);n_{CH_4}=\dfrac{16}{16}=1\left(mol\right)\)
\(\overline{M}_A=\dfrac{16+16}{0,25+1}=25,6\left(g/mol\right)\)
- Xét B:
Do \(V_{Cl_2}=V_{O_2}\Rightarrow n_{Cl_2}=n_{O_2}\)
Giả sử \(n_{Cl_2}=n_{O_2}=1\left(mol\right)\)
\(\overline{M}_B=\dfrac{1.71+1.32}{1+1}=51,5\left(g/mol\right)\)
\(d_{A/B}=\dfrac{25,6}{51,5}\approx0,497\)
\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)
\(\overline{M_x}=24.2=48\)
\(\left\{{}\begin{matrix}SO_2:64\\O_2:32\end{matrix}\right.\) 48 = \(\dfrac{16}{16}=1\)
\(\Rightarrow n_{SO_2=}n_{O_2}=0,3mol\)
1. \(m_{hh}=0,3.64+0,3.32=28,8g\)
2. \(\%V_{SO_2}=\dfrac{0,3.22,4}{13,44}.100\%=50\%\)
\(\Rightarrow\%V_{O_2}=50\%\)
3. \(m_{SO_2}=0,3.64=19,2g\)
\(m_{O_2}=0,3.32=9,6g\)
a) \(\left\{{}\begin{matrix}n_{Cl_2}+n_{O_2}=\dfrac{6,72}{22,4}=0,3\\\overline{M}=\dfrac{71.n_{Cl_2}+32.n_{O_2}}{n_{Cl_2}+n_{O_2}}=2.29=58\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{Cl_2}=0,2\left(mol\right)\\n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{Cl_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{O_2}=\dfrac{0,1}{0,3}.100\%=33,33\%\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m_{Cl_2}=0,2.71=14,2\left(g\right)\\m_{O_2}=0,1.32=3,2\left(g\right)\end{matrix}\right.\)
Áp dụng quy tắc đường chéo:
\(a.\\ \Rightarrow\dfrac{V_{Cl_2}}{V_{O_2}}=\dfrac{15,6}{23,4}=\dfrac{2}{3}\\ \Rightarrow\left\{{}\begin{matrix}\%V_{Cl_2}=40\%\\\%V_{O_2}=60\%\end{matrix}\right.\)
\(b.\)
Ta có: \(\dfrac{n_{Cl_2}}{n_{O_2}}=\dfrac{2}{3}\Leftrightarrow\dfrac{m_{Cl_2}}{m_{O_2}}=\dfrac{71.2}{32.3}=\dfrac{71}{48}\Leftrightarrow48m_{Cl_2}-71m_{O_2}=0\)
Mặt khác: \(m_{Cl_2}+m_{O_2}=5,95\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cl_2}=3,55\left(g\right)\\m_{O_2}=2,4\left(g\right)\end{matrix}\right.\)
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