\(M_X=2,5.16=40\)(g/mol)
\(\rightarrow\dfrac{V_{CO_2}}{V_{O_2}}=\dfrac{n_{CO_2}}{n_{O_2}}=\dfrac{40-32}{44-40}=2\)
Mà \(V_{CO_2}+V_{O_2}=30\left(L\right)\)
\(\rightarrow V_{CO_2}=20\left(L\right);V_{O_2}=10\left(L\right)\)
\(\rightarrow M_Y=\dfrac{20.44+10.32+32V}{V+20+10}=2,25.16=36\)
\(\rightarrow V=30\left(L\right)\)