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 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l) At 1.00 atm and 23 oC, what is the volume of carbon dioxide formed by the combustion of 3.4 g of butane?

5.638 L is the volume formed

Explanation:

The reaction for the combustion is:

2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)

First of all we convert the mass of butane to moles (mass / molar mass)

3.4 g / 58 g/mol = 0.058 moles

As we assume, the oxygen in excess ratio between butane and carbon dioxide is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of carbon dioxide

0.058 moles of butane must produce (0.058 . 8) /2 = 0.234 moles of CO₂

Now we apply the Ideal Gases Law to find out the volume formed.

P . V = n . R . T

1atm . V = 0.234 mol . 0.082 L.atm/mol.K . 296K

V = 0.234 mol . 0.082 L.atm/mol.K . 296K / 1atm = 5.68 L

false

because it does not have the potential to be entirely used up

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current