Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 15.5 L of butane (C4H10): 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)

Answer : The number of moles of oxygen gas required are, 4.49 moles.

Explanation :

The balanced chemical reaction is:

First we have to calculate the moles of butane gas.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 22.4 L volume of butane gas present in 1 mole of butane gas

So, 15.5 L volume of butane gas present in mole of butane gas

Now we have to calculate the moles of oxygen gas.

From the balanced chemical reaction we conclude that,

As, 2 moles of butane gas react with 13 moles of oxygen gas

So, 0.692 moles of butane gas react with moles of oxygen gas

Therefore, the number of moles of oxygen gas required are, 4.49 moles.