Question

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 1

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

Explanation :

The balanced chemical reaction is:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(l)$

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 $\frac{15.5}{22.4}=0.692$ 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 $\frac{13}{2}\times 0.692=4.49$ moles of oxygen gas

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