Question

Select each of the reactions and observe the reactants taking part in the reactions. Suppose you are carrying out each of these reactions starting with five moles of each reactant. Observe that some reactants are in excess whereas some reactants limit the amount of products formed.
Classify each of the reactants as a limiting reactant or an excess reactant for a reaction starting with five moles of each reactant.

A. H2 in formation of water
B. O2 in formation of water
C. CH4 in combustion of methane
D. N2 in formation of ammonia
E. H2 in formation of ammonia
F. O2 in combustion of methane

Answer 1

Answer:

A. In the formation of water, H2 is the limiting reactant and O2 is the excess reactant.

B. In the Combustion of methane (CH4), O2 is the limiting reactant and CH4 is the excess reactant.

C. In the formation of ammonia (NH3), H2 is the limiting reactant and N2 is the excess reactant.

Explanation:

The question suggests that each reactant has 5 moles.

A. Formation of water. Water is produced when H2 and O2 combine together according to the balanced equation below:

2H2 + O2 —> 2H2O

From the balanced equation above,

2 moles of H2 reacted with 1 mole of O2.

Therefore, 5 moles of H2 will react with = 5/2 = 2.5 moles of O2.

From the above calculations, we can see that there are left over for O2 as only 2.5 moles reacted out of the 5 moles that was given.

Therefore, H2 is the limiting reactant and O2 is the excess reactant.

B. Combustion of methane. Combustion is simply a reaction in the presence of oxygen. The balance equation for the Combustion of methane (CH4) is given below:

CH4 + 2O2 —> CO2 + 2H2O

From the balanced equation above,

1 mole of CH4 reacted with 2 moles of O2.

Therefore, 5 moles of CH4 will react with = 5 x 2 = 10 moles of O2.

From the calculations made above, we can see that it requires higher amount of O2 to react with 5 moles CH4. Therefore, O2 is the limiting reactant and CH4 is the excess reactant.

C. Formation of ammonia.

Ammonia is obtained when N2 and H2 combine according to the balanced equation below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 5 moles of N2 will react with = 5 x 3 = 15 moles of H2.

From the calculations made above, a higher amount of H2 is required to react with 5 moles of N2. Therefore, H2 is the limiting reactant and N2 is the excess reactant.