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 answer is barium and copper bromide

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.