19.4 Stoichiometric calculations
Quantitative aspects of chemical change
19.3 Amount of substance
1 Theoretical yield
Exercise 19.11
19.4 Stoichiometric calculations (ESAGF)
Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is important to know how much product will be formed in a chemical reaction, or how much of a reactant is needed to make a specific product.
The following diagram shows how the concepts that we have learnt in this chapter relate to each other and to the balanced chemical equation:
Worked example 15: Stoichiometric calculation 1
What volume of oxygen at S.T.P. is needed for the complete combustion of \(\text{2}\) \(\text{dm$^{3}$}\) of propane (\(\text{C}_{3}\text{H}_{8}\))? (Hint: \(\text{CO}_{2}\) and \(\text{H}_{2}\text{O}\) are the products in this reaction (and in all combustion reactions))
Write the balanced equation
\[\text{C}_{3}\text{H}_{8}\text{(g)} + 5\text{O}_{2}\text{(g)} \rightarrow 3\text{CO}_{2}\text{(g)} + 4\text{H}_{2}\text{O (g)}\]
Find the ratio
Because all the reactants are gases, we can use the mole ratios to do a comparison. From the balanced equation, the ratio of oxygen to propane in the reactants is \(5:1\).
Find the answer
One volume of propane needs five volumes of oxygen, therefore \(\text{2}\) \(\text{dm$^{3}$}\) of propane will need \(\text{10}\) \(\text{dm$^{3}$}\) of oxygen for the reaction to proceed to completion.
What mass of iron (II) sulphide is formed when \(\text{5,6}\) \(\text{g}\) of iron is completely reacted with sulphur?
\[\text{Fe (s)} + \text{S (s)} \rightarrow \text{FeS (s)}\]
Calculate the number of moles
We find the number of moles of the given substance:
\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{5,6}\text{ g}}{\text{55,8}\text{ g·mol$^{-1}$}} \\ & = \text{0,1}\text{ mol} \end{align*}
Find the mole ratio
We find the mole ratio between what was given and what you are looking for. From the equation \(\text{1}\) \(\text{mol}\) of Fe gives \(\text{1}\) \(\text{mol}\) of \(\text{FeS}\). Therefore, \(\text{0,1}\) \(\text{mol}\) of iron in the reactants will give \(\text{0,1}\) \(\text{mol}\) of iron sulphide in the product.
Find the mass of iron sulphide
\begin{align*} m & = nM \\ & = (\text{0,1}\text{ mol})(\text{87,9}\text{ g·mol$^{-1}$}) \\ & = \text{8,79}\text{ g} \end{align*}
The mass of iron (II) sulphide that is produced during this reaction is \(\text{8,79}\) \(\text{g}\).
Theoretical yield (ESAGG)
When we are given a known mass of a reactant and are asked to work out how much product is formed, we are working out the theoretical yield of the reaction. In the laboratory, chemists almost never get this amount of product. In each step of a reaction a small amount of product and reactants is "lost" either because a reactant did not completely react or some other unwanted products are formed. This amount of product that you actually got is called the actual yield. You can calculate the percentage yield with the following equation:
\[\% \text{ yield } = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\]
Worked example 17: Industrial reaction to produce fertiliser
Sulphuric acid (\(\text{H}_{2}\text{SO}_{4}\)) reacts with ammonia (\(\text{NH}_{3}\)) to produce the fertiliser ammonium sulphate (\((\text{NH}_{4})_{2}\text{SO}_{4}\)). What is the theoretical yield of ammonium sulphate that can be obtained from \(\text{2,0}\) \(\text{kg}\) of sulphuric acid? It is found that \(\text{2,2}\) \(\text{kg}\) of fertiliser is formed. Calculate the % yield.
\[\text{H}_{2}\text{SO}_{4}\text{(aq)} + 2\text{NH}_{3}\text{(g)} \rightarrow \text{(NH}_{4}\text{)}_{2}\text{SO}_{4}\text{(aq)}\]
Calculate the number of moles of the given substance
\begin{align*} n_{\text{H}_{2}\text{SO}_{4}} & = \frac{m}{M} \\ & = \frac{\text{2 000}\text{ g}}{\text{98,12}\text{ g·mol$^{-1}$}} \\ & = \text{20,38320...}\text{ mol} \end{align*}
From the balanced equation, the mole ratio of \(\text{H}_{2}\text{SO}_{4}\) in the reactants to \((\text{NH}_{4})_{2}\text{SO}_{4}\) in the product is \(1:1\). Therefore, \(\text{20,3830...}\) \(\text{mol}\) of \(\text{H}_{2}\text{SO}_{4}\) forms \(\text{20,3830...}\) \(\text{mol}\) of \((\text{NH}_{4})_{2}\text{SO}_{4}\).
Write the answer
The maximum mass of ammonium sulphate that can be produced is calculated as follows:
\begin{align*} m & = nM \\ & = (\text{20,3830...}\text{ mol})(\text{132,18}\text{ g·mol$^{-1}$}) \\ & = \text{2 694,25193...}\text{ g} \end{align*}
The maximum amount of ammonium sulphate that can be produced is \(\text{2,694}\) \(\text{kg}\).
Calculate the \(\%\) yield
\begin{align*} \% \text{ yield } & = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \\ & = \frac{\text{2,2}}{\text{2,694}} \times 100 \\ & = \text{81,66}\% \end{align*}
Worked example 18: Calculating the mass of reactants and products
Barium chloride and sulphuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.
\[\text{BaCl}_{2} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{BaSO}_{4} + 2\text{HCl}\]
If you have \(\text{2}\) \(\text{g}\) of \(\text{BaCl}_{2}\):
What quantity (in g) of \(\text{H}_{2}\text{SO}_{4}\) will you need for the reaction so that all the barium chloride is used up?
What mass of \(\text{HCl}\) is produced during the reaction?
Find the number of moles of barium chloride
\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{2}\text{ g}}{\text{208,2}\text{ g·mol$^{-1}$}} \\ & = \text{0,0096}\text{ mol} \end{align*}
Find the number of moles of sulphuric acid
According to the balanced equation, 1 mole of \(\text{BaCl}_{2}\) will react with 1 mole of \(\text{H}_{2}\text{SO}_{4}\). Therefore, if \(\text{0,0096}\) \(\text{mol}\) of \(\text{BaCl}_{2}\) react, then there must be the same number of moles of \(\text{H}_{2}\text{SO}_{4}\) that react because their mole ratio is \(1:1\).
Find the mass of sulphuric acid
\begin{align*} m & = nM \\ & = (\text{0,0096}\text{ mol})(\text{98,12}\text{ g·mol$^{-1}$}) \\ & = \text{0,94}\text{ g} \end{align*}
(answer to 1)
Find the moles of hydrochloric acid
According to the balanced equation, 2 moles of \(\text{HCl}\) are produced for every 1 mole of the two reactants. Therefore the number of moles of \(\text{HCl}\) produced is \(2(\text{0,0096})\), which equals \(\text{0,0192}\) \(\text{mol}\).
Find the mass of hydrochloric acid
\begin{align*} m & = nM \\ & = (\text{0,0192}\text{ mol})(\text{36,46}\text{ g·mol$^{-1}$}) \\ & = \text{0,7}\text{ g} \end{align*}
Textbook Exercise 19.7
Diborane, \(\text{B}_{2}\text{H}_{6}\), was once considered for use as a rocket fuel. The combustion reaction for diborane is:
\[\text{B}_{2}\text{H}_{6}\text{(g)} + 3\text{O}_{2}\text{(g)} \rightarrow 2\text{HBO}_{2}\text{(g)} + 2\text{H}_{2}\text{O (l)}\]
If we react \(\text{2,37}\) \(\text{g}\) of diborane, how many grams of water would we expect to produce?
Solution not yet available
Sodium azide is a commonly used compound in airbags. When triggered, it has the following reaction:
\[2\text{NaN}_{3}\text{(s)} \rightarrow 2\text{Na (s)} + 3\text{N}_{2}\text{(g)}\]
If \(\text{23,4}\) \(\text{g}\) of sodium azide is used, how many moles of nitrogen gas would we expect to produce? What volume would this nitrogen gas occupy at STP?
Photosynthesis is a chemical reaction that is vital to the existence of life on Earth. During photosynthesis, plants and bacteria convert carbon dioxide gas, liquid water, and light into glucose (\(\text{C}_{6}\text{H}_{12}\text{O}_{6}\)) and oxygen gas.
Write down the equation for the photosynthesis reaction.
Balance the equation.
If \(\text{3}\) \(\text{mol}\) of carbon dioxide are used up in the photosynthesis reaction, what mass of glucose will be produced?