If you start with 128.25 grams of nonane,C9H20, how many grams of oxygen are required for this reaction?

Answer 1

Firstly, write a balanced equation of the reaction. When nonane is burnt in oxygen, it produces carbon dioxide and water.

C9H20 +14O2 -> 9CO2 + 10H2O

Now, you must find out how many moles of nonane there is. The formula for finding moles is n=m/M, which is mass over relative molecular mass.
The relative molecular mass of nonane is
Mr = (9x12.01)+(20x1.008)
=128.25
You know the mass is 128.25
Therefore, n=128.25/128.25
=1mol
Now, you know that for a reaction to take place, for every 1mol of nonane, 14mol of oxygen is required. We HAVE 1 mol of nonane, so we need 14 mol of oxygen.

Now, we use the same formula as before to find the mass of oxygen. The relative molecular mass of oxygen is 16.
Therefore, 14=m/16
m=14x16
=224g

Therefore, the amount of oxygen required for the reaction is 224g.