1) Convert 12.9 liters of Oxygen to mol at the given conditions:
PV = nRT ⇒ n = PV/RT
n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]
n = 0.636 mol of O2
2) use the stoichiometry derived from the balanced chemical equation
1mol C2H4 / 3 mol O2 = x mol C2H4 / 0.636 mol O2
x = 0.636 / 3 mol O2 = 0.212 mol O2.
Answer: 0.212 mol O2
Answer:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
Explanation:
Chemical equation:
CO + O₂ → CO₂
Balanced chemical equation:
2CO + O₂ → 2CO₂
The standard enthalpy for the formation of CO = -110.5 kj/mol
The standard enthalpy for the formation of O₂ = 0 kj/mol
The standard enthalpy for the formation of CO₂ = -393.5 kj/mol
Now we will put the values in equation:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]
ΔH0reaction = -283 kj/mol
