In the unbalanced chemical reaction for the combustion of propane determine at standard temperature and pressure how many liters of carbon dioxide gas are produced if 15 liters of oxygen gas are completely consumed

9 liters of CO₂ are produced by this combustion

Explanation:

In order to determine the volume of produced CO₂, we start with the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

We need, O₂ density to find out the mass, that has reacted.

δ O₂ = O₂ mass / O₂ volume  → δ O₂  . O₂ volume = O₂ mass

δ O₂ = 1.429 g /dm₃ (1dm³ = 1L) 1.429 g/L  . 15L = 21.4 g of O₂

We convert the mass to moles: 21.4 g . 1mol / 32 g = 0.670 moles

By stoichiometry, 5 moles of O₂ can produce 3 moles of CO₂

Then, 0.670 moles of O₂ will produce (0.670 . 3) /5 = 0.402 moles of dioxide.

We apply Ideal Gases Law for STP, to find out the CO₂ volume

V = (n . R  . T) / P → V = (0.402 mol . 0.082 . 273K) / 1 atm = 8.99 L ≅ 9 L

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.