Consider the reaction for the combustion of butane (c4h10), a component of liquefied petroleum (lp) gas. 2c4h10(g)+13o2(g)→8co2(g)+10h2o(g)< br /> 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?

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                                

you gotta add the following molecules so we can give u an answer

explanation: