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Limiting Reactants

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Limiting Reactants Revision

Limiting Reactants

In a lot of reactions in which two reactants are used, it is common to use more of one of the reactants then is actually needed. When this is the case, this reaction is said to be in excess. This is make sure that we use up all of the other reactant. The reactant that is not in excess is known as the limiting reactant (also known as the limiting reagent). The limiting reactant is so called as it limits the amount of product that can be formed. The amount of product formed will be directly proportional to the amount of limiting reactant used. For example, if you were to triple the amount of limiting reactant used then the amount of product formed would also triple (provided the limiting reactant was still limiting).Once the limiting reactant has been used up, the reaction stops.

Calculations with the limiting reactant

Knowing which of our reactants is limiting allows us to calculate the mass of product formed. For example, if we add 5\text{ g} sodium metal \left(\text{Na}\right) to an excess of water, we can calculate the mass of sodium hydroxide \left(\text{NaOH}\right) formed using only the initial mass of sodium added (and a periodic table so we can calculate the relative masses). We start with the balanced equation for the reaction:

2\text{Na} + 2\text{H}_2\text{O} \rarr 2\text{NaOH} + \text{H}_2

We can see from this equation that the ratio between \text{Na} and \text{NaOH} is 1:1. The number of moles of \text{Na} that react with the water is equal to that of the \text{NaOH} produced. 

Next, we calculate the number of moles of \text{Na} in a 5\text{ g} sample:

\text{Moles Na} = \frac{5}{23} =\underline{0.217 \text{ mol}}

As the ratio of \text{Na} to \text{NaOH} is 1:1, we know that the number of sodium moles equals the number of sodium hydroxide moles:

\text{Moles Na} =\text{Moles NaOH}

\text{Moles NaOH} = \underline{0.217 \text{ mol}}

With the moles of \text{NaOH} in hand, all that remains is to use its formula mass to calculate the mass formed:

\text{Mass NaOH} = 0.217 \times 40 = \underline{8.70 \text{ g}}

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Example 1: Calculating Masses

A student carries out an experiment to determine how much energy is released by respiration. To do this they burn a \textcolor{#00bfa8}{3.2\text{ g}} sample of glucose \left(\text{C}_6\text{H}_{12}\text{O}_6, \text{M}_r =\textcolor{#f21cc2}{180}\right) in an excess of oxygen. During the experiment, the following reaction takes place:

\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rarr 6\text{CO}_2 + 6\text{H}_2\text{O}

Calculate the mass of \text{CO}_2  \left(\text{M}_r =\textcolor{#327399}{44}\right) produced:

[3 marks]

Step 1: Calculate the moles of glucose reacted.

\text{Moles C}_6\text{H}_{12}\text{O}_6 \text{ reacted} = \frac{\textcolor{#00bfa8}{3.2}}{\textcolor{#f21cc2}{180}} = \textcolor{#008d65}{0.018 \text{ mol}}

Step 2: Find the moles of \text{CO}_2 produced from the equation.

\text{Ratio} = 1:6

\begin{aligned}\text{Moles CO}_2 &= 6\times \text{Moles C}_6\text{H}_{12}\text{O}_6\\ & =\textcolor{#008d65}{0.11\text{ mol}}\end{aligned}

Step 3: Calculate the mass of \text{CO}_2 produced.

\text{Mass CO}_2 = 0.11\times \textcolor{#327399}{44} = \textcolor{#008d65}{4.8 \text{ g}}

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Example 2: Calculating Masses and Determining the Limiting Reactant

\textcolor{#00bfa8}{6.02\text{ g}} of Rubidium Hydroxide \left(\text{RbOH}, \text{M}_r =\textcolor{#f21cc2}{102}\right) was reacted with \textcolor{#327399}{8.3\text{ g}} of phosphoric acid \left(\text{H}_3\text{PO}_4, \text{M}_r =\textcolor{#a233ff}{98}\right) to form rubidium phosphate \left(\text{Rb}_3\text{PO}_4, \text{M}_r =\textcolor{#eb6517}{350}\right) and water in the following reaction:

3\text{RbOH} + \text{H}_3\text{PO}_4 \rarr \text{Rb}_3\text{PO}_4 + 3\text{H}_2\text{O}

A. Deduce which reactant is the limiting reactant. 

[3 marks]

Step A 1: Calculate the moles of \text{RbOH}.

\text{Moles of RbOH} =\frac{\textcolor{#00bfa8}{6.02}}{\textcolor{#f21cc2}{102}}=\textcolor{#008d65}{0.059\text{ mol}}

Step A 2: Calculate the moles of \text{H}_3\text{PO}_4.

\text{Moles H}_3\text{PO}_4=\frac{\textcolor{#327399}{8.3}}{\textcolor{#a233ff}{98}}=\textcolor{#008d65}{0.085\text{ mol}}

Step A 3: Determine the limiting reactant.

\text{Moles of RbOH} < \text{Moles H}_3\text{PO}_4

RbOH is the limiting reagent. 

B. Calculate the mass of rubidium phosphate formed.

[2 marks]

Step B 1: Find the ratio of \text{RbOH} to \text{Rb}_3\text{PO}_4 from the equation.

\text{RbOH}:\text{Rb}_3\text{PO}_4=3:1

\text{Moles Rb}_3\text{PO}_4 = \frac{\text{Moles RbOH}}{3}=\textcolor{#008d65}{0.020\text{ mol}}

Step B 2: Calculate the mass of rubidium phosphate produced.

\text{Mass Rb}_3\text{PO}_4=0.020 \times \textcolor{#eb6517}{350} = \textcolor{#008d65}{7.0\text{ g}}

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Limiting Reactants Example Questions

The limiting reactant is the reactant in a reaction that is not in excess. 

The amount of product formed is directly proportional to the amount of limiting reactant used.

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Calculation should be broken down into steps (one mark for each correct calculation).

Step 1: Calculate the moles of \text{SO}_3 reacted

\text{Moles of SO}_3 = \frac{23}{80} = \underline{0.288\text{ mol}}

Step 2: Calculate the moles of \text{H}_2\text{SO}_4

\text{Moles of SO}_3 =\text{Moles of H}_2\text{SO}_4

Step 3: Calculate the mass of \text{H}_2\text{SO}_4 formed.

\text{Mass of H}_2\text{SO}_4 = 98 \times 0.288 = \underline{28.22\text{ g}}

 

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Calculation should be broken down into steps (one mark for each correct calculation).

Step 1: Calculate the moles of \text{HCl}.

\text{Moles HCl} =\frac{6.33}{36} =\underline{0.176\text{ mol}}

Step 2: Calculate the moles of \text{NaOH}.

\text{Moles NaOH} =\frac{8.99}{40} =\underline{0.225\text{ mol}}

Step 3: Determine which reactant is in excess.

\text{Moles HCl} < \text{Moles NaOH}

NaOH is in excess 

 

 

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