Question

How many grams of O2 are required to burn 18 grams of C5H12? C5H12 + 8O2  5CO2 + 6H2O

Answer 1

Molar mass :

O2 = 31.99 g/mol

C5H12 = 72.14 g/mol

C5H12 + 8 O2 = 5 CO2 + 6 H2O

74.14 g ----------------- 8 x 31.99 g
18 g -------------------- ? ( mass of O2)

18 x 8 x 31.99 / 74.14 =

4606.56 / 74.14 => 62.133 g of O2

hope this helps!

Answer 2

Answer:

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