Hướng dẫn giải

- Áp dụng ĐLBT khối lượng ta có: $\begin{align}& {{\text{m}}_{\text{C}{{\text{H}}_{\text{4}}}}}\text{+}{{\text{m}}_{{{\text{O}}_{\text{2}}}}}\text{=}{{\text{m}}_{\text{C}{{\text{O}}_{\text{2}}}}}\text{+}{{\text{m}}_{{{\text{H}}_{\text{2}}}\text{O}}} \\& \Rightarrow {{\text{m}}_{\text{C}{{\text{H}}_{\text{4}}}}}\text{=m=}{{\text{m}}_{\text{C}{{\text{O}}_{\text{2}}}}}\text{+}{{\text{m}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{- }{{\text{m}}_{{{\text{O}}_{\text{2}}}}}\text{=1,4+1,6-0,4=2,6 gam} \\\end{align}$

-  Chọn đáp án B.