Question

Use standard heat of formations to find the heat for the combustion of 2.6 g of acetylene gas C2H2 to form
carbon dioxide and water vapor

Answer 1

we have:

Hof(C2H2(g)) = 226.73 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(CO2(g)) = -393.509 KJ/mol

Hof(H2O(g)) = -241.818 KJ/mol

we have the Balanced chemical equation as:

2 C2H2(g) + 5 O2(g) ---> 4 CO2(g) + 2 H2O(g)

deltaHo rxn = 4*Hof(CO2(g)) + 2*Hof(H2O(g)) - 2*Hof( C2H2(g)) - 5*Hof(O2(g))

deltaHo rxn = 4*(-393.509) + 2*(-241.818) - 2*(226.73) - 5*(0.0)

deltaHo rxn = -2511.132 KJ/mol

Molar mass of C2H2 = 2*MM(C) + 2*MM(H)

= 2*12.01 + 2*1.008

= 26.036 g/mol

mass of C2H2 = 2.6 g

we have below equation to be used:

number of mol of C2H2,

n = mass of C2H2/molar mass of C2H2

=(2.6 g)/(26.036 g/mol)

= 9.986*10^-2 mol

delta Ho = deltaHorxn * number of mol

= -2511.132 KJ/mol * 9.986*10^-2 mol

= -250.8 KJ

Answer: -250.8 KJ