$n_{$CO_{2}}$ = $\frac{8}{22.4}$ = $\frac{5}{14}$ (mol)
$C_{6}$$H_{12}$$O_{6}$ + $O_{2}$ → $2C_{2}$$H_{5}$$OH_{}$ + $2CO_{2}$ (xt: men rượu, $t^{o}$ ≈ $30^{o}$C - $35^{o}$C)
Theo PTHH, ta có:
+) $n_{C_{2}H_{5}OH}$ = $n_{CO_{2}}$ = $\frac{5}{14}$ mol
+) $n_{C_{6}H_{12}O_{6}}$ = $0.5n_{C_{2}H_{5}OH}$ = $\frac{5}{28}$ mol
⇒ $m_{C_{6}H_{12}O_{6}}$/H=100% = $\frac{5}{28}$ × 180 ≈ 32.143 (g)
a) $m_{C_{2}H_{5}OH}$ = $\frac{5}{14}$ × 46 ≈ 16.43 (g)
b) $m_{C_{6}H_{12}O_{6}}$/H=75% = 32.143 ÷ 75% ≈ 42.86 (g)
c) $m_{C_{6}H_{12}O_{6}}$/H=65% = 32.143 ÷ 65% ≈ 49.45 (g)
~ GỬI BẠN ~
