Question

if 90 litres of butane is reacted with 700 litres of oxygen calculate the composition of resulting misture

Answer 1

Given:

The volume of butane = 90 L

The volume of oxygen = 700 L

To Find:

The composition of the resulting mixture.

Calculation:

- No of moles of butane in 90 L = 90/22.4 = 4.02

- The reaction of butane and oxygen is given as:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

- 1 mole of butane reacts with 6.5 moles of oxygen to produce 4 moles of carbon dioxide and 5 moles of water.

⇒ The moles of oxygen required to react with 4.02 mole of butane = 6.5 × 4.02 = 26.13

⇒ The amount of oxygen used = 26.13 × 22.4 = 585.31 L

- The amount of oxygen left = 700 - 585.31 = 114.69 L

- The moles of carbon dioxide produced = 4 × 4.02 = 16.08

⇒ The amount of carbon dioxide produced = 16.08 × 22.4 = 360.19 L

- The moles of water produced = 5 × 4.02 = 20.1

⇒ The amount of water produced = 20.1 × 22.4 = 450.24 L

- So, the resulting mixture contains 114.69 L of Oxygen, 360.19 L of Carbon Dioxide and 450.24 L of Water.