Giả sử có 5 mol không khí, 1 mol SO2
=> \(\left\{{}\begin{matrix}n_{O_2}=1\left(mol\right)\\n_{N_2}=4\left(mol\right)\end{matrix}\right.\)
X \(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{1}{1+1+4}.100\%=16,67\%\\\%V_{O_2}=\dfrac{1}{1+1+4}.100\%=16,67\%\\\%V_{N_2}=100\%-16,67\%-16,67\%=66,66\%\end{matrix}\right.\)
\(\overline{M}_X=\dfrac{32.1+28.4+64.1}{1+4+1}=\dfrac{104}{3}\left(g/mol\right)\)
mX = mY = 208 (g)
Gọi số mol SO2 pư là a (mol)
PTHH: 2SO2 + O2 --to--> 2SO3
Trc pư: 1 1 0
Pư: a--->0,5a------->a
Sau pư: (1-a) (1-0,5a) a
=> \(n_{khí\left(sau.pư\right)}=\left(1-a\right)+\left(1-0,5a\right)+a+4=6-0,5a\left(mol\right)\)
\(\overline{M}_Y=\dfrac{\overline{M}_X}{0,93}=\dfrac{10400}{279}\left(g/mol\right)\)
=> \(\dfrac{208}{6-0,5a}=\dfrac{10400}{279}\)
=> a = 0,84 (mol)
Y gồm \(\left\{{}\begin{matrix}SO_2:0,16\left(mol\right)\\O_2:0,58\left(mol\right)\\SO_3:0,84\left(mol\right)\\N_2:4\left(mol\right)\end{matrix}\right.\)
nkhí = 5,58 (mol)
=> \(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,16}{5,58}.100\%=2,867\%\\\%V_{O_2}=\dfrac{0,58}{5,58}.100\%=10,394\%\\\%V_{SO_3}=\dfrac{0,84}{5,58}.100\%=15,054\%\\\%V_{N_2}=\dfrac{4}{5,58}.100\%=71,685\%\end{matrix}\right.\)
