$n_{H_{2}}$ = $\frac{2.24}{22.4}$ = 0.1 (mol)
$2CH_{3}$$OH_{}$ + $2Na_{}$ → $2CH_{3}$$ONa_{}$ + $H_{2}$ (1)
$C_{2}$$H_{4}$$(OH)_{2}$ + $2Na_{}$ → $C_{2}$$H_{4}$$(ONa)_{2}$ + $H_{2}$ (2)
$2C_{3}$$H_{5}$$(OH)_{3}$ + $6Na_{}$ → $2C_{3}$$H_{5}$$(ONa)_{3}$ + $3H_{2}$ (3)
Theo PTHH (1), (2) và (3) có:
$n_{OH/X}$ = $n_{H/H_{2}}$ = 2$n_{H_{2}}$ = 0.1 × 2 = 0.2 (mol)
Lại có: $n_{C/X}$ = $n_{OH/X}$ ⇒ $n_{C/X}$ = 0.2 (mol)
⇒ Trong 2m (g) X có 0.4 mol C
$2CH_{4}$$O_{}$ + $3O_{2}$ → $2CO_{2}$ + $4H_{2}$$O_{}$
$2C_{2}$$H_{6}$$O_{2}$ + $5O_{2}$ → $4CO_{2}$ + $6H_{2}$$O_{}$
$2C_{3}$$H_{8}$$O_{3}$ + $7O_{2}$ → $6CO_{2}$ + $8H_{2}$$O_{}$
Bảo toàn C ⇒ $n_{CO_{2}}$ = $n_{C/A}$ = 0.4 (mol)
$V_{CO_{2}(đktc)}$ = 0.4 × 22.4 = 8.96 (l)
~ GỬI BẠN ~
