Calculate the mass of water produced when 4.95 g of butane reacts with excess oxygen.

7.676 g  of water

The combustion of a hydrocarbon yields water and carbon dioxide.

The complete combustion of butane is given by the equation;

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

We are given 4.95 g C₄H₁₀

We need to calculate the mass of water produced.

Number of moles = Mass ÷ Molar mass

Molar mass = 58.12 g/mol

Number of moles = 4.95 g ÷ 58.12 g/mol

                            = 0.0852 moles

From the equation;

2 moles of butane completely burns in air to produce 10 moles of water

Therefore, 0.0852 moles of butane will produce;

= 0.0852 moles × 10/2

= 0.426 moles

Mass is given by multiplying number of moles of a compound by its molar mass.

Thus;

Mass of water = 0.426 moles × 18.02 g/mol

                       = 7.676 g

Therefore, 7.676 g of water are produced when 4.95 g of butane completely burns in excess oxygen.

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.