Đáp án:
a) C4H10
b) 42g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
nC{O_2} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol = > nC = 0,6\,mol\\
n{H_2}O = \dfrac{{13,5}}{{18}} = 0,75\,mol = > nH = 1,5\,mol\\
mC + mH = 0,6 \times 12 + 1,5 = 8,7g\\
= > B:C,H\\
nC:nH = 0,6:1,5 = 2:5\\
= > CTDGN:{C_2}{H_5}\\
MB = 58g/mol = > 29n = 58 = > n = 2\\
= > CTPT:{C_4}{H_{10}}\\
CTCT:C{H_3} - C{H_2} - C{H_2} - C{H_3}\\
C{H_3} - CH(C{H_3}) - C{H_3}\\
b)\\
nNaOH = 0,25 \times 2 = 0,5\,mol\\
\dfrac{{nNaOH}}{{nC{O_2}}} = \dfrac{{0,5}}{{0,6}} = 0,833\\
NaOH + C{O_2} \to NaHC{O_3}\\
nNaHC{O_3} = nNaOH = 0,5\,mol\\
mNaHC{O_3} = 0,5 \times 84 = 42g
\end{array}\)