a) MA = 13.2 = 26(g/mol)
\(m_C=\dfrac{26.92,3}{100}=24\left(g\right)=>n_C=\dfrac{24}{12}=2\left(mol\right)\)
\(m_H=26-24=2\left(g\right)=>n_H=\dfrac{2}{1}=2\left(mol\right)\)
=> CTHH: C2H2
b) \(n_A=\dfrac{67,2}{22,4}=3\left(mol\right)\)
mC = 3.2.12 = 72 (g)
mH = 3.2.1 = 6(g)