How many grams of o2 are required to burn 18 grams of c5h12?
c5h12 + 8o2  5co2 + 6h2o

63,86 g O2  

Explanation:

You will first find the molecular weight (MW) of C5H12 and O2 with help of a periodic table.

Then C= 12,011 , H=1,008  y O=15,999      

So:  

C5H12 = 12,011*5 + 1,008*12      

MW =72,151 g/ mol C5H12      

O2 = 15,999*2      

MW = 31,998 g/ mol O2    

Use the equation to know that 1 mol C5H12 equals 8 mol O2    

Now you will start with the data of 18  grams of C5H12 and you obtain:      

18 g C5H12 x 1 mol C5H12     x 8 mol O2    x   31,998 g O2  = 63,86 g O2                                                    .                     72,151 g C5H12    1 mol C5H12      1 mol O2  

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.