a,
\(n_C=n_{CO2}=0,6\left(mol\right)\)
\(n_H=2n_{H2O}=1,5\left(mol\right)\Rightarrow m_H=1,5\left(g\right)\)
mC+ mH= mB \(\Rightarrow\) B là hidrocacbon
\(n_C:m_H=0,6:1,5=2:5\)
Nên CTĐGN (C2H5)n
\(M=58\Rightarrow n=2\)
Vậy CTPT C4H10
B ko phân nhánh nên có CTCT CH3-CH2-CH2-CH3
b,
\(n_{H2O}=0,75\left(mol\right)\)
\(n_{CO2}=0,6\left(mol\right)\)
\(n_{NaOH}=0,5\left(mol\right)=n_{OH^-}\)
\(\frac{n_{OH^-}}{n_{CO2}}=0,83\)
\(\Rightarrow\) Chỉ tạo muối NaHCO3
\(NaOH+CO_2\rightarrow NaHCO_3\)
\(\Rightarrow\) nCO2 dư
\(n_{NaHCO3}=n_{NaOH}=0,5\left(mol\right)\)
\(\Rightarrow m_{NaHCO3}=42\left(g\right)\)
