Ankan X : \(C_nH_{2n+2}\)
\(n_{CO_2} = \dfrac{2,24}{22,4} = 0,1(mol)\)
Bảo toàn nguyên tố với C:
\(n_X = \dfrac{n_{CO_2}}{n} = \dfrac{0,1}{n}mol\\ \Rightarrow m_X = \dfrac{0,1}{n}.(14n+2) = 1,6\ gam\\ \Rightarrow n = 1\)
Vậy X : CH4(metan)
nCO2 = 2.24/22.4 = 0.1 (mol)
BT cacbon :
nCnH2n+2 = nCO2/n = 0.1/n (mol)
M = 1.6/0.1/n = 16n
=> 14n + 2 = 16n
=> 2n = 2
=> n = 1
CT : CH4 ( metan)