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

it is b that is the answer.

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.