Bài 3.1
C2H2 + $\frac{5}{2}$ O2 → 2CO2 + H2O
$n_{C2H2}$ = $\frac{6,72}{22,4}$ = 0,3 mol
$n_{O2}$ = $n_{C2H2}$×$\frac{5}{2}$ = 0,75 mol
$V_{O2}$ = 0,75×22,4= 16,8l
$n_{CO2}$ = $n_{C2H2}$×2 = 0,6mol
$m_{CO2}$ = 0,6×44= 26,4g
Bài 3.2
C2H4+Br2→C2H4Br2
$n_{C2H4}$ = $\frac{3,36}{22,4}$ = 0,15 mol
$n_{Br2}$ = $n_{C2H4}$ = 0,15mol
$V_{Br2}$ =$\frac{0,15}{1}$ = 0,15l = 150ml
$n_{C2H4Br2}$=$n_{C2H4}$ = 0,15mol
$m_{C2H4Br2}$ = 0,15×188= 28,2g