Câu 4:D
$k=\dfrac{4a}{a}+3=7$
⇒$\text{CTTQ của X:}C_nH_{2n-12}O_6$
$C_nH_{2n-12}O_6+\dfrac{3n-12}{2}O_2\xrightarrow{}nCO_2+(n-6)H_2O$
$n_{CO_2}-n_{H_2O}=a.(7-1)⇒n_{CO_2}=6a+b(1)$
⇒$V_{CO_2}=22.4(6a+b)⇒D$
Câu 5:D
$k=\dfrac{0,1}{a}+3$
$a=\dfrac{n_{CO_2}-n_{H_2O}}{k-1}=\dfrac{2,75-2,55}{\dfrac{0,1}{x}+3-1}=0,05(mol)$
⇒$k=5$
BTNT O: $n_{O_2}=\dfrac{2,75.2+2,55-0,05.6}{2}=3,875(mol)$
⇒BTKL:$m_{X}=2,75.44+2,55.18-3,875.32=42,9(g)$
$\text{Gọi công thức của X là:}(RCOO)_3C_3H_5$
$(RCOO)_3C_3H_5+3NaOH\xrightarrow{t^o}3RCOONa+C_3H_5(OH)_3$
0,05 : 0,15 : 0,15 : 0,05 (mol)
BTKL: $m_{muối}=42,9+0,15.40-92.0,05=44,3(g)$
=>D
