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

you need to give more information

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.