Question

Answer 1

\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=\dfrac{28}{5}=5,6\left(l\right)\Rightarrow n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ Vì:\dfrac{0,2}{1}>\dfrac{0,25}{2}\Rightarrow CH_4dư\)