Question

Tripalmitin can be used in oil lamps and its combustion is described by the balanced equation:

2 C₅₁H₉₈O₆(l) ₊ 151O₂(g) _--> 102 CO₂(g) + 98 H₂O(l)

mol weight: 807.34 32.00 44.01 18.02

a).Calculate the mass of carbon dioxide in kilograms that would be produced by the complete combustion of 1 gallon of tripalmitin.

b). Is 6.50 kg of O₂ enough to result in complete combustion of 1 gallon of tripalmitin? Yes or No... Explain.

Answer 1

as 1 gallon = 3.79 Kg

molecular mass of triplamitin C51H98O6 =807.34g/mol

molecular mass of O2 =32 g/mol

molecular mass of CO2 = 44.01 g/mol

molecular mass of H2O = 18. 02g/mol

moles of triplamitin = given mass /molecular weight = 3.79kg/807.34g/mol = 4.69 *10^(-3) Kmol

1 Kmol = 1000 mol

(a) let mass of CO2 = x Kg

moles of CO2 = x/44.01 Kmol

as in given reaction,

2 C51H98O6 + 151O2 - - - > 102 CO2 + 98 H2O

it is clear that

2 moles of triplamitin = produces 102 moles CO2

then 1 moles of triplamtin =produces 102/2 moles of CO2

now 4.69*10^(-3) kmoles of triplamitin =

produces 102 *4.69*10^(-3) /2 = 0.239 kmoles of CO2.

kmoles of CO2 = 0.239 = x/44.01 kmol

then x(kg) =0.239 *44.01 =10.54 kg

thus 10.54 kg CO2 is produced.

(b) now as in reaction

moles of triplamitin for 1 gallon = 4.69 *10^(-3)Kmol

2 moles of triplamitin =reacts with 151 moles of O2

1 mole of triplamitin = reacts with 151/2 moles of O2

then 4.69*10^(-3) Kmole of triplamitin =

reacts with 4.69 *10^(-3) *151/2 =0.354 Kmol

now Kmoles of O2 =

mass(Kg) /molecular mass of O2(g/mol)

mass (Kg) = 0.354 *32 = 11.33

So 11.33 Kg of O2 is required for complete combustion. So (b) statement is false.

Answer in NO.