Question

If the acetylene tank contains 37.0 mol of C2H2 and the oxygen tank contains 81.0 mol of O2, what is the limiting reactant for this reaction?

Answer 1

Answer: $O_2$ is considered as the limiting reagent for this reaction.

Explanation:

Limiting reagent is the reagent which limits the formation of the product.

Excess reagent is the reagent which is present in excess in a chemical reaction.

For the combustion of acetylene, the reaction follows:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2+2H_2O$

By Stoichiometry,

5 moles of oxygen gas reacts with 2 moles of acetylene.

So, 81 moles of oxygen gas will react with = $\frac{2}{5}\times 81$ = 32.4 moles of acetylene.

As, the required amount of acetylene is less than the given amount. So, it is considered as an excess reagent and oxygen gas is the limiting reagent.

Answer 2

Answer:

Oxygen is the limiting reactant of this reaction.

Explanation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

2 moles of acetylene reacts with 5 moles of oxygen.

Then 37 moles of acetylene will react with :

$\frac{5}{2}\times 37.0 mol=92.5 moles$ of oxygen gas.

But only 81 moles of oxygen gas is present.

Then 81.0 moles of oxygen gas will react with:

$\frac{2}{5}\times 81.0 mol=32.4 moles$ of acetylene.

And still acetylene will remain after.

So, from the above calculation we cans see that oxygen is present in limiting amount which will limit the reaction.Hence, the limiting reactant of this reaction is oxygen.