Limiting Reagents
A limiting reagent is a reactant that is totally consumed in a chemical reaction. The limiting reagent, as its name implies, limits the amount of product produced during the reaction. For example, let's take a look at the following reaction in which hydrogen and oxygen react to form water: \[2\text{H}_2(g)+\text{O}_2(g)\rightarrow2\text{H}_2\text{O}(l).\] When this reaction occurs, exactly two hydrogen molecules and an oxygen molecule react to produce two water molecules. Suppose there are 10 moles of hydrogen and 7 moles of oxygen. Since the molar ratio of hydrogen and oxygen molecules is not 2:1, one must be left over after the reaction has fully occurred, which would be oxygen in this case. Thus, all 10 moles of hydrogen will react with 5 moles of oxygen to produce 10 moles of water, leaving 2 moles of oxygen, as shown in the equation below: \[\begin{align} &2\text{H}_2(g)&&+&&\text{O}_2(g)&&\rightarrow&&2\text{H}_2\text{O}(l)\\ &\ \ \ \ 10&& &&\ \ \ \ 7&& &&\ \ \ \ \ 0\\ &\ -10&& &&\ -5&& &&+10\\ \hline &\ \ \ \ \ 0&& &&\ \ \ \ 2&& &&\ \ \ \ 10. \end{align}\] In this example, hydrogen is the limiting reagent and oxygen is the excess reagent. The amount of product formed is limited by the amount of hydrogen. In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents.
The following is the chemical equation of the combustion of methane. 64 grams of methane reacts with 96 grams of oxygen. Identify the limiting reagent, and calculate the amount (in grams) of carbon dioxide produced from this reaction. Use the following atomic weights: \(\text{H}=1, \text{C}=12, \text{O}=16.\)
\[\text{CH}_4+2\text{O}_2\rightarrow\text{CO}_2+2\text{H}_2\text{O}.\]
First, we find the number of moles of reactants we have. Since one mole of methane is \(12+4\times1=16\) grams and one mole of oxygen is \(2\times16=32\) grams, we have \(64\div16=4\) moles of methane and \(96\div32=3\) moles of oxygen. Since one methane molecule is to react with two oxygen molecules, oxygen is the limiting reagent and methane is the excess reagent. 1.5 moles of methane will react with 3 moles of oxygen to produce 1.5 moles of carbon dioxide and 3 moles of water, as shown in the equation below: \[\begin{align} &\text{CH}_4&&+&&2\text{O}_2&&\rightarrow&&\text{CO}_2&&+&&2\text{H}_2\text{O}\\ &\ \ 4&& &&\ \ 3&& &&\ \ 0&& &&\ \ \ 0\\ &-1.5&& &&-3&& &&+1.5&& &&+3\\ \hline &\ 2.5&& &&\ \ 0&& &&\ 1.5&& &&\ \ \ 3. \end{align}\] Therefore, the limiting reagent is oxygen, and the amount of produced carbon dioxide is \(1.5\times(12+2\times16)=66\) grams. \(_\square\)