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gia su co 1 mol SO2 ,suy ra co 5 mol khong khi tuc la co 1 mol O2 va 4 mol N2
ti khoi cua A la d(A)= ( 1*64+1*32+4*28)/(1+5)=208/6
khi nung hon hop A voi V2O5 xay ra phan ung
2SO2 + O2 ----> 2SO3 (1)
2a ---->a -------->2a
dat so mol oxi phan ung la a suy ra so mol SO2 bang so mol SO3 = 2a
sau phan ung (1) so mol cua hon hop giam di a mol -> so mol cua hon hop B la (6-a) mol, khoi luong cua B = khoi luong cua A = 208 gam -> d(B) = 208/(6-a)
d(A)/d(B) =(6-a)/6 = 0.93 -> a= 0.42 -> so mol SO2 = 2a = 0.84 mol
trong hon hop A do oxi du nen hieu suat phan ung tinh theo SO2
H= 0.84/1 = 0.84 = 84% ->dap an C
\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)
\(\overline{M_x}=24.2=48\)
\(\left\{{}\begin{matrix}SO_2:64\\O_2:32\end{matrix}\right.\) 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. \(m_{SO_2}=0,3.64=19,2g\)
\(m_{O_2}=0,3.32=9,6g\)
Có \(A\left\{{}\begin{matrix}n_{SO_2}+n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\\dfrac{64.n_{SO_2}+32.n_{O_2}}{n_{SO_2}+n_{O_2}}=25,6.2=51,2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{SO_2}=0,3\left(mol\right)\\n_{O_2}=0,2\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,3}{0,5}.100\%=60\%\\\%V_{O_2}=\dfrac{0,2}{0,5}.100\%=40\%\end{matrix}\right.\)
Gọi số mol SO2 phản ứng là x (mol)
PTHH: 2SO2 + O2 --> 2SO3
Trc pư: 0,3 0,2 0
Pư: x------>0,5x------>x
Sau pư: (0,3-x) (0,2-0,5x) x
=> \(M_B=\dfrac{m_B}{n_B}=\dfrac{m_A}{n_B}=\dfrac{25,6}{\left(0,3-x\right)+\left(0,2-0,5x\right)+x}=32.2=64\)
=> x = 0,2
=> \(B\left\{{}\begin{matrix}SO_2:0,1\left(mol\right)\\O_2:0,1\left(mol\right)\\SO_3:0,2\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,1}{0,1+0,1+0,2}.100\%=25\%\\\%V_{O_2}=\dfrac{0,1}{0,1+0,1+0,2}.100\%=25\%\\\%V_{SO_3}=\dfrac{0,2}{0,1+0,1+0,2}.100\%=50\%\end{matrix}\right.\)
- Xét hỗn hợp khí A:
Gọi x,y lần lượt là số mol của SO2 và O2 trong hỗn hợp. (x,y>0) (mol)
\(x+y=\dfrac{11,2}{22,4}=0,5\left(1\right)\\ Mà:M_A=25,6.M_{H_2}=25,6.2=51,2\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow\dfrac{64x+32y}{0,5}=51,2\\ \Leftrightarrow64x+32y=25,6\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}x+y=0,5\\64x+32y=25,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\\ \Rightarrow\%V_{\dfrac{SO_2}{A}}=\dfrac{0,3}{0,5}.100=60\%\Rightarrow\%V_{\dfrac{O_2}{A}}=100\%-60\%=40\%\)
- Xét hỗn hợp khí B:
Gọi a là số mol SO3 được tạo thành trong hhB (mol) (a,b>0)
\(PTHH:2SO_2+O_2\rightarrow\left(xt,t^o\right)2SO_3\\ \Rightarrow n_{SO_2\left(hhB\right)}=0,3-a\left(mol\right)\\ n_{O_2\left(hhB\right)}=0,2-0,5a\left(mol\right)\\ M_{hhB}=32.M_{H_2}=32.2=64\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow\dfrac{80a+\left(0,2-0,5a\right).32+\left(0,3-a\right).64}{a+\left(0,2-0,5a\right)+\left(0,3-a\right)}=64\\ \Leftrightarrow a=0,2\\ \Rightarrow hhB\left\{{}\begin{matrix}SO_3:0,2\left(mol\right)\\SO_2:0,1\left(mol\right)\\O_2:0,1\left(mol\right)\end{matrix}\right.\\ \Rightarrow\%V_{\dfrac{SO_3}{hhB}}=\dfrac{0,2}{0,2+0,1+0,1}.100=50\%\\ \%V_{\dfrac{SO_2}{hhB}}=\%V_{\dfrac{O_2}{hhB}}=\dfrac{0,1}{0,2+0,1+0,1}.100=25\%\)
Em xem có gì không hiểu thì hỏi lại nhá!
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: \(m_{Cl_2}+m_{O_2}=5,95\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cl_2}=3,55\left(g\right)\\m_{O_2}=2,4\left(g\right)\end{matrix}\right.\)
