If 156.06 g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed? C3H8 + O2 → CO2 + H2O Select one: a. 255.0 b. 765.1 c. 191.3 d. 31.88

The correct option is;

a. 255.0 g

Explanation:

The given information are;

Mass of propane, C₃H₈ in the combustion reaction = 156.06 g

The equation of the combustion reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

From the balanced chemical equation of the reaction, we have;

One mole of propane, C₃H₈ reacts with five moles oxygen gas, O₂, to form three moles of carbon dioxide, CO₂, and  four moles of water, H₂O

The molar mass of propane gas = 44.1 g/mol

The number of moles, n, of propane gas = Mass of propane gas/(Molar mass of propane gas) = 156.06/44.1 = 3.54 moles

Given that one mole of propane gas produces 4 moles of water molecule (steam) H₂O, 3.54 moles of propane gas will produce 4×3.54 = 14.16 moles of  (steam) H₂O

The mass of one mole of H₂O = 18.01528 g/mol

The mass of 14.16 moles of H₂O = 14.16 × 18.01528 = 255.0 g

The mass of H₂O produced = 255.0 g

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.