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

Answer:

The answer to your question is:

a) 3.87 mol of CO2

b) 14.63 mol of O2

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