Question

a) If 13.8g of butane gas reacts with 21.4g of oxygen gas, what will be the theoretical yield of each product produced in the reaction (assume no side reactions)?

b) Which reactant is in excess and by how much?

c) Type of reaction

d) Redox or Non-Redox

e) What is the theoretical change in H for this reaction in kcal/mol?

Answer 1

a)

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

116.24 g             416 g                                 352.1 g           180.2 g

13.8 g                 21.4 g                                  ??

116.24 g C4H10   --------------> 416 g O2

13.8 g C4H10    -------------->   ??

mass of O2 needed = 13.8 x 416 / 116.24 = 49.4 g

but we have 21.4 g O2. so limiting reagent is O2 .

416 g O2   -------------> 352.1 g CO2   and 180.2 g H2O

21.4 g O2   ------------>   ??

mass of CO2 produced = 21.4 x 352.1 / 416

mass of CO2 produced = 18.1 g

mass of H2O produced = 9.27 g

b)

butane is excess .

mass of excess butane = 13.8 - 5.98

                                         = 7.82 g

c)

combustion reaction

d)

redox