General chemistry

Question

Butane (C4H10) reacts with oxygen to form water and carbon dioxide. What mass of oxygen is needed to burn 23.15 grams of butane? How many water molecules would this reaction produce?

Answer 1

First thing you need to do is write out a "balanced" chemical equation in order to know how many mols of each type of molecule is involved in the reaction.

C4H10 + O2 --> H2O + CO2

Normally I would first balance it out based on C followed by H then O

C4H10 + (13/2)O₂ --> 5H₂O + 4CO₂

2C₄H₁₀ + 13O₂ --> 10H₂O + 8 CO₂ (Balanced chemical equation)

Now knowing the mass of butane you have use its molecular weight (g/mol) to determine the mols of butane:

23.15 g * (1 mol/58.14 g) = 0.39818 mols butane

Looking at the balanced chemical equation we know that 2 mols of butane are needed for every 13 mols of O₂. Use this relationship to determine the mols of O₂:

0.39818 mols butane * (13 mols O₂ / 2 mols C₄H₁₀) = 2.588 mols O₂

Use the molecular weight of oxygen to determine the mass needed:

2.588 mols * (32.0 g /mol) = 82.82 g O₂

To find out # of water molecules produced use the relationship between water and butane (mols wise) to determine mols of H2O then use Avogadro's number to determine # molecules of H2O

0.39818 mols butane * (10 mols H2O / 2 mols butane) = 1.991 mols H2O

Avogadro's # : 6.022 * 10²³ (something) / mol

1.991 mols H2O * (6.022 * 10^23 molecules / mol) = 1.20 * 10^24 H2O molecules