Gọi \(\left\{{}\begin{matrix}n_{O_2\left(bđ\right)}=a\left(mol\right)\\n_{SO_2\left(bđ\right)}=b\left(mol\right)\end{matrix}\right.\)
Ta có: \(\dfrac{32a+64b}{a+b}=24.2=48(g/mol)\)
=> a = b
Giả sử a = b = 1 (mol)
=> mkhí sau pư = 1.32 + 1.64 = 96 (g)
Gọi số mol O2 pư là x (mol)
PTHH: \(2SO_2+O_2\underrightarrow{t^o,V_2O_5}2SO_3\)
Bđ 1 1
Pư: 2x<-----x---------->2x
Sau pư: (1-2x) (1-x) 2x
Ta có: \(\dfrac{96}{\left(1-2x\right)+\left(1-x\right)+2x}=30.2=60\left(g/mol\right)\)
=> x = 0,4 (mol)
Sau pư thu được \(\left\{{}\begin{matrix}SO_2:0,2\left(mol\right)\\O_2:0,6\left(mol\right)\\SO_3:0,8\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,2}{0,2+0,6+0,8}.100\%=12,5\%\\\%V_{O_2}=\dfrac{0,6}{0,2+0,6+0,8}.100\%=37,5\%\\\%V_{SO_3}=\dfrac{0,8}{0,2+0,6+0,8}.100\%=50\%\end{matrix}\right.\)
\(\%O_2\left(pư\right)=\dfrac{0,4}{1}.100\%=40\%\)
\(\%SO_2\left(pư\right)=\dfrac{0,8}{1}.100\%=80\%\)
