Question

3) 320 grams of methane react with 400 grams of oxygen gas. Limiting Reactant?

CH4 + 2O2 CO2 + 2H2O

Answer 1

Answer:

Explanation:

We must start by creating a balanced reaction between Methane and Oxygen. As methane is a hydrocarbon reacting with oxygen this will be a combustion reaction resulting in carbon dioxide and water.

The combustion reaction is:

C

H

4

+

2

O

2

→

C

O

2

+

2

H

2

O

Now we want to find how many moles of each reactant we have to find which one is limiting.

If we do a rough calculation for finding how many moles of  

C

H

4

and how many moles of  

O

2

we have we get the following:

Using the molarity equation:

n

=

m

M

M=molar mass of the compound/element,

m=mass in grams of the compound/element

n=number of moles of the compound/element

n(Oxygen)=

32

32

=

1

mol

Oxygen is our limiting reactant

=

2

mol

However, in the reaction equation, we need 2 moles of oxygen for every one mole of Methane, and we only have 1 mole of oxygen and 2 moles of Methane. That means methane is in excess and oxygen is thus our limiting reactant.

So in this reaction we would only be able to react 1 mol of  

O 2  with 0.5 mol of  C H 4

.