Đáp án B
\(n_{CO_2}=\dfrac{3,3}{44}=0,075\left(mol\right)\)
\(\rightarrow n_C=n_{CO_2}=0,075\left(mol\right)\)
\(n_{H_2O}=\dfrac{4,5}{18}=0,25\left(mol\right)\)
\(\rightarrow n_H=2.n_{H_2O}=0,25.2=0,5\left(mol\right)\)
\(m=m_C+m_H=12.0,075+1.0,5=1,4\left(g\right)\)
