BTKL :
\(m_A=m_{C_3H_8}=8.8\left(g\right)\)
\(n_{C_3H_8}=\dfrac{8.8}{44}=0.2\left(mol\right)\)
\(\Rightarrow n_{C_3H_8\left(pư\right)}=0.2\cdot90\%=0.18\left(mol\right)\)
\(C_3H_8\underrightarrow{t^0,xt}C_aH_{2a+2}+C_bH_{2b}\left(a+b=3\right)\)
\(n_{C_3H_8\left(dư\right)}=0.02\left(mol\right)\)
\(M_A=\dfrac{8.8}{0.18\cdot2+0.02}=23.16\left(\dfrac{g}{mol}\right)\)
\(C_3H_8 \to H_2 + C_3H_6\\ C_3H_8 \to CH_4 + C_2H_4\\ C_3H_8 \to C_3H_{8\ dư}\\ m_A = m_{propan}= 8,8(gam)\\ n_{C_3H_8} = \dfrac{8,8}{44} = 0,2\\ n_{C_3H_8\ dư} = 0,2 - 0,2.90\% =0,02(mol)\\ n_A = 2n_{C_3H_8\ pư} + n_{C_3H_8\ dư} = 0,2.90\%.2 + 0,02=0,38(mol)\\ M_A = \dfrac{8,8}{0,38} = 23,167\)