What volume of oxygen is consumed when 9g of water is formed when butane is burnt?
To determine the volume of oxygen consumed when 9 grams of water is formed during the combustion of butane, we need to follow these steps:
1. Write a balanced chemical equation for the combustion of butane (C₄H₁₀):
C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O
The equation shows that for every 1 mole of butane burned, we need 13/2 moles of oxygen to produce 5 moles of water.
2. Calculate the moles of water formed:
The molar mass of water (H₂O) is 18 g/mol. Therefore, 9 grams of water is equal to 9 / 18 = 0.5 moles.
3. Determine the moles of oxygen required:
Using the stoichiometry of the balanced equation, we can see that for every 5 moles of water, we need 13/2 moles of oxygen.
(0.5 moles of water) x (13/2 moles of O₂ / 5 moles of H₂O) = 0.65 moles of O₂
4. Calculate the volume of oxygen gas:
The ideal gas law, PV = nRT, can be used to calculate the volume (V) of a gas.
Given that the temperature (T) and pressure (P) are constant, we only need to calculate the number of moles (n) of oxygen consumed.
The molar volume of an ideal gas at standard conditions (STP) is 22.4 L/mol.
Therefore, the volume of oxygen consumed would be:
(0.65 moles of O₂) x (22.4 L/mol) = 14.56 L
So, when 9 grams of water is formed during the combustion of butane, approximately 14.56 liters of oxygen gas would be consumed.