Question

Butane, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete combustion of butane is 2c4h10(g)+13o2(g)→8co2(g)+10h2o(l) at 1.00 atm and 23 ∘c, what is the volume of carbon dioxide formed by the combustion of 1.60 g of butane?

Answer 1

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=$\frac{1.6g}{58.124\frac{g}{mol}}$

= 0.027 mole of butane

0.027 mole of butane =$\frac{8}{2 }\times 0.027$

                                     = 0.108 mole of CO₂.

Using ideal gas equation,  

$P\times V=n\times R\times T$

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,

$V=\frac{0.018\times 0.0821\times 296}{1}$

V= 2.62 L

So the volume will be 2.62 L