Limiting reagent

Chemical equations specify precise amounts of reactants that will combine to give products. In a real chemical reaction, arbitrary amounts of reactants are mixed together and allowed to form products, which means generally that one of the reactants will be used up completely, while excess amounts of other reactants will remain after the reaction has run to completion. The species that is fully consumed is called the limiting reagent. The limiting reagent will then determine what the yield of product will be.

When the quantities of reactants that are allowed to combine are specified, identifying the limiting reagent amounts to determining which of the reacting species yields the smallest amount of product.

Consider the following example. Sulfuric acid is produced in the reaction:

$$2{\rm SO}_2 + {\rm O}_2 + 2{\rm H}_2{\rm O} \longrightarrow 2{\rm H}_2{\rm SO}_4$$

Suppose 400 g of SO$_2$, 175 g of O$_2$ and 125 g of H$_2$O are mixed and allows to react. Which reactant is the limiting reagent?

Solution: First convert the given masses to moles:

$${\rm moles\ of\ }{\rm SO}_2 = {400 {\rm\ g} \over 64.06 {\rm\ g/mol}} = 6.24 {\rm\ mol\ }{\rm SO}_2$$

$${\rm moles\ of\ }{\rm O}_2 = {175 {\rm\ g} \over 31.996 {\rm\ g/mol}} = 5.47 {\rm\ mol\ }{\rm O}_2$$

$${\rm moles\ of\ }{\rm H}_2{\rm O} = {125 {\rm\ g} \over 18.02 {\rm\ g/mol}} = 6.94 {\rm\ mol\ }{\rm H}_2{\rm O}$$

From the reaction equation, we see that

2 moles of SO$_2$ yields 2 moles of H$_2$SO$_4$

or

1 mole of SO$_2$ yields 1 mole of H$_2$SO$_4$

or

6.24 moles of SO$_2$ yields 6.24 moles of H$_2$SO$_4$

Similarly,

1 mole of O$_2$ yields 2 moles of H$_2$SO$_4$

or

5.47 mole of O$_2$ yields 10.94 moles of H$_2$SO$_4$

and

2 moles of H$_2$O yields 2 moles of H$_2$SO$_4$

or

1 mole of H$_2$O yields 1 mole of H$_2$SO$_4$

or

6.94 moles of H$_2$O yields 6.94 moles of H$_2$SO$_4$

Since SO$_2$ yields the smallest quantity of product, it is the limiting reagent.