(Stoichiometry)

Question

1 C₅H₁₂(g) + 8 O₂(g) → 5 CO₂(g) + __ H₂O(g) when balancing the above equation, how many moles of water are produced when 1 mole of C₅H₁₂ reacts with excess Oxygen Gas? 

Answer 1

Hi, Mohamed,

To solve this question, we first need to balance the equation (which describes the combustion of pentane, and conversion to carbon dioxide and water):

1C₅H₁₂(g) + 8 O₂(g) → 5 CO₂(g) + __ H₂O(g)

The simplest way to do this is work from left to right to verify that each element is represented equally on the left and then right side of the reaction arrow. In this case, we see that 1 molecule/mole of pentane represents five carbon atoms (or moles of carbon). This is balanced with the production of 5 molecules/moles of carbon dioxide on the right side. For hydrogen, we see that 1 molecule of pentane accounts for 12 atoms of hydrogen. On the right side, the only hydrogen product shown is water (with 2 hydrogen atoms per 1 water molecule). In this case, to balance 12 hydrogen atoms on the left, we need 6 water molecules as 6 X 2 is 12. Finally, there are 8 X 2 or 16 oxygen atoms on the reactant side of the equation - with 5 X 2 (10) oxygen atoms accounted for as carbon dioxide, and 6 oxygen atoms as part of the 6 water molecules, the oxygen atom count is balanced. The balanced equation, then, is:

1C₅H₁₂(g) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(g)

To now answer the question, when 1 mole of C₅H₁₂ reacts with excess oxygen, it will produce 6 moles of water.

'Hope that helps,

Chris

Answer 2

the starting point here is balancing the chemical equation so that you can find the mole ratio between C₅H₁₂ and water. for this reaction that ends up being

C5​H12​(g) + 8O2​(g) → 5CO2​(g) + 6H2​O(g)

meaning that the molar ratio between C₅H₁₂ and water is 1:6.

therefore, when one mole of C₅H₁₂ reacts with excess oxygen gas, 6 moles of water are produced.