\(n_{CO_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)
\(n_{H_2O}=\dfrac{9.9}{18}=0.55\left(mol\right)\)
\(BảotoànO:\)
\(n_{O_2}=n_{CO_2}+\dfrac{1}{2}n_{H_2O}=0.35+\dfrac{0.55}{2}=0.625\left(mol\right)\)
\(V_{O_2}=0.625\cdot22.4=14\left(l\right)\)
\(\Rightarrow A\)
