Question

What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the reaction to be as follows: C4H10(g)+132O2(g)→4CO2(g)+5H2O(g),ΔHrxn=−2658 kJ

Answer 1

32.8 g of Butane is required and 99.3 g of CO₂ is produced

Explanation:

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }$    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

                                        = 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced