Question

When 3g of a mixture of 1-butene (C H)butane (CH10) and an inert gas was burned in excess of oxygen
8.80g of Co, and 4.14g ofH Owere obtained. What is percentage by mass of butane in the mixture.​

Answer 1

Answer:

96.67%

Explanation:

the balanced equation for this reaction is:

C4H10 + (13/2)O2 --> 4CO2 + 5H2O

inert gases don't react so do not matter in the reaction

moles: mass/mr                                  n = moles

n(CO2) = 8.8/(12+(2 x 16)

            = 0.2

we will now use ratio of CO2 : C4H10 which is 4:1

n(C4H10) = 0.2/4

               = 0.05

mass of C4H10 needed to make 0.2 moles of CO2 = 0.05 x [(4 x 12) + 10]

                                                                                      = 2.9g

percentage mass = (2.9/3) x 100

                             = 96.67%

mass = n x mr