$n_{CaCO_{3}}$ = $\frac{14.4}{100}$ = 0.144 (mol)
$CO_{2}$ + $Ca(OH)_{2}$ → $CaCO_{3}↓$
$C_{2}$$H_{6}$$O_{}$ + $3O_{2}$ → $2CO_{2}$ + $3H_{2}$$O_{}$ (đk: $t^{o}$ cao)
Theo PTHH, ta có: $n_{C_{2}H_{6}O_{}}$ = $0.5n_{CO_{2}}$ = $0.5n_{CaCO_{3}}$ = 0.144 × 0.5 = 0.072 (mol)
$m_{C_{2}H_{6}O_{}}$ = 0.072 × 46 = 3.312 (g)
$V_{C_{2}H_{6}O_{}}$ = $\frac{3.312}{0.8}$ = 4.14 (ml)
Độ rượu = $\frac{4.14}{4.5}$ × 100 = $92^{o}$
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