Question

Consider the reaction for the combustion of butane (C4H10), a component of liquefied petroleum (LP) gas. 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)

How many moles of CO2 are produced when 0.968 mol of butane are burned?

How many moles of oxygen are required to burn 2.25 mol of butane?

Answer 1

The answer to your question is:

a) 3.87 mol of CO2

b) 14.63 mol of O2

Step-by-step explanation:

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

a)

CO2 = ?

Butane 0.968 moles

From the balance equation 2 moles of butane ---------- 8 moles of CO2

0.968 moles of butane ------ x

x = (0.968 x 8) / 2

x = 3.87 moles of CO2

b)

moles O2 = ?

butane = 2.25 mol

From the balance equation

2 moles of butane ---------- 13 mol O2

2.25 mol ---------- x

x = (2.25 x 13) / 2

x = 14.63 mol of O2