Respuesta :
Reaction of combustion of butane is a as follows:
2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O:
As it can be seen from the equation that 2 mole of butane produce= 8 mole of butane produce
1 mole of butane produce= 4 mole of butane produce
1.60 g of butane=[tex]\frac{1.6g}{58.124\frac{g}{mol}}[/tex]
= 0.027 mole of butane
0.027 mole of butane =[tex]\frac{8}{2 }\times 0.027[/tex]
= 0.108 mole of CO₂.
Using ideal gas equation,
[tex]P\times V=n\times R\times T[/tex]
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=1 atm
T=23+273 K = 296 K
R=0.0821 atm L mol ⁻¹
Number of moles of gas, n= 0.018
Putting all the values in the above equation,
[tex]V=\frac{0.018\times 0.0821\times 296}{1}[/tex]
V= 2.62 L
So the volume will be 2.62 L
