$🎿dangduccan🎿$

$V_{H_{2}O}=1250-550=700ml$

$V_{CO_{2}}=550-250=300ml$

$V_{N_{2}}=100ml$

$2NH_{3}$ $+\frac{3}{2}$ $O_{2}$ $→$ $N_{2}+$ $3H_{2}O$

$\text{Ta có:}$ $V_{NH_{3}}=$ $2V_{N_{2}}$ $=100.2=200ml$

$→$ $V_{H_{2}O\text{đốt NH3}}=$$1,5V_{NH_{3}}=300ml$

$→$ $V_{H_{2}O\text{đốt hidrocacbon}}=700-300=400ml$

$\text{Đốt hidrocacbon thu được }$$V_{CO_{2}}<$$V_{H_{2}O}$ $\text{nên hidrocacbon là ankan}$

$\text{Đặt CTTQ hidrocacbon là}$ $C_{n}H_{2n+2}$

$C_{n}H_{2n+2}$ $+$ $\frac{3n+1}{2}O_{2}$ $→$ $nCO_{2}$ $+$ $(n+1 )H_{2}O$

$\text{Ta có:}$ $\frac{V_{CO_{2}}}{V_{H_{2}O}}$ $=$ $\frac{n_{CO_{2}}}{n_{H_{2}O}}$ $=$ $\frac{n}{n+1}$$=$ $\frac{300}{400}$ $=$$\frac{3}{4}$

$→$ $3(n+1)=4n$

$→$ $n=3$

$\text{Vậy CTPT hidrocacbon là }$$C_{3}H_{8}$