Question

Consider the combustion reaction for acetylene. 2C2H2(l) + 5O2(g) mc001-1.jpg 4CO2(g) + 2H2O(g) 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

First we need to ensure that the reaction is balanced

$2C_2H_2_(_l_) + 5O_2_(_g_) ==> 4CO_2_(_g_) + 2H_2O_(_g_)$

It is balanced, so we use stochiometry to figure the amount of product formed by each individual reactant. For instance, we find $CO_2$ by each reactant.

$CO_2$ from $C_2H_2$

$molCO_2= 37mol C_2H_2 \times \frac{4molCO_2}{2molC_2H_2} = 74 mol CO_2$

$CO_2$ from $O_2$

$mol CO_2=81 mol O_2 \times\frac{3molCO_2}{5molO_2} = 64.8 mol CO_2$

The oxygen $(O_2)$ will be finished when 64.8mol of $(CO_2)$ have been produced. There will be excess acetylene $C_2H_2$ at the end of the reaction. Hence Oxygen is the limiting reactant.

Answer 2

Answer: Oxygen

Explanation:

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

As can be seen from the balanced chemical equation,  2 moles of acetylene react with 5 moles of oxygen.

Thus 37 moles of acetylene will react with $=\frac{5}{2}\times 37=92.5moles$ of oxygen.

But $O_2$ available is only 81.0 moles.

Thus now we take it the other way:

5 moles of oxygen  react with 2 moles of acetylene

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

Limiting reagent is the reagent which limits the formation of product. Excess reagent is one which is in excess and thus remains unreacted.

Thus oxygen is the limiting reagent and acetylene is the excess reagent as (37-32.4)= 4.6 moles of acetylene are left.