a, PTHH: $C_{3}H_{8}$ + 5$O_{2}$ $\xrightarrow{t^o}$ 3$CO_{2}$ + 4$H_{2}O$ 

b, Ta có: $n_{C_{3}H_{8}}$ = $\frac{V_{C_{3}H_{8}}}{24,79}$ = $\frac{49,58}{24,79}$ $=$ $2$ $(mol)$

Thep $pt$: $n_{CO_{2}}$ $=$ 3×$n_{C_{3}H_{8}}$ $=$ $3×2$ $=$ $6$ $(mol)$

⇒ $V_{CO_{2}(đkc)}$ = $24,79$×$n_{CO_{2}}$ = $24,79$×$6$ = $148.74$ $(l)$

Thep $pt$: $n_{O_{2}}$ $=$ 5×$n_{C_{3}H_{8}}$ $=$ $5×2$ $=$ $10$ $(mol)$

⇒ $V_{O_{2}(đkc)}$ = $24,79$×$n_{O_{2}}$ = $24,79$×$10$ = $247,9$ $(l)$

$@NhuQuynhQ.T$ $gửi$ $ạ$

a) Phương trình hóa học của phản ứng:
\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]

b) Tính toán:
\[ \text{Số mol C}_3\text{H}_8 = \frac{49,58}{22,4} \approx 2,213 \, \text{mol} \]
\[ \text{Số mol CO}_2 = 2,213 \times 3 = 6,639 \, \text{mol} \]
\[ \text{Thể tích CO}_2 = 6,639 \times 22,4 = 148,7 \, \text{lít} \]
\[ \text{Số mol O}_2 = 2,213 \times 5 = 11,065 \, \text{mol} \]
\[ \text{Thể tích O}_2 = 11,065 \times 22,4 = 247,0 \, \text{lít} \]