Stoichiometric calculations
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:
Example 1: Stoichiometric calculation 1
Question
What volume of oxygen at S.T.P. is needed for the complete combustion of 2 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))
Answer
Write the balanced equation
${\text{C}}_{3}{\text{H}}_{8}\text{(g)}+5{\text{O}}_{2}\text{(g)}\to 3\text{C}{\text{O}}_{2}\text{(g)}+4{\text{H}}_{2}\text{O}\text{(g)}$
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 2 dm^{3} of propane will need 10 dm^{3} of oxygen for the reaction to proceed to completion.
Example 2: Stoichiometric calculation 2
Question
What mass of iron (II) sulphide is formed when 5,6 g of iron is completely reacted with sulphur?
Answer
Write the balanced equation
$\text{Fe}\text{(s)}+\text{S}\text{(s)}\to \text{FeS}\text{(s)}$
Calculate the number of moles
We find the number of moles of the given substance:
Find the mole ratio
We find the mole ratio between what was given and what you are looking for. From the equation 1 mol of Fe gives 1 mol of $\text{FeS}$. Therefore, 0,1 mol of iron in the reactants will give 0,1 mol of iron sulphide in the product.
Find the mass of iron sulphide
The mass of iron (II) sulphide that is produced during this reaction is 8,79 g.
Theoretical yield
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:
Example 3: Industrial reaction to produce fertiliser
Question
Sulphuric acid (${\text{H}}_{2}{\text{SO}}_{4}$) reacts with ammonia (${\text{NH}}_{3}$) to produce the fertiliser ammonium sulphate (${\left({\text{NH}}_{4}\right)}_{2}{\text{SO}}_{4}$). What is the theoretical yield of ammonium sulphate that can be obtained from 2,0 kg of sulphuric acid? It is found that 2,2 kg of fertiliser is formed. Calculate the % yield.
Answer
Write the balanced equation
${\text{H}}_{2}{\text{SO}}_{4}\text{(aq)}+2{\text{NH}}_{3}\text{(g)}\to {\left({\text{NH}}_{4}\right)}_{2}{\text{SO}}_{4}\text{(aq)}$
Calculate the number of moles of the given substance
Find the mole ratio
From the balanced equation, the mole ratio of ${\text{H}}_{2}{\text{SO}}_{4}$ in the reactants to ${\left({\text{NH}}_{4}\right)}_{2}{\text{SO}}_{4}$ in the product is $1:1$. Therefore, 20,383 mol of ${\text{H}}_{2}{\text{SO}}_{4}$ forms 20,383 mol of ${\left({\text{NH}}_{4}\right)}_{2}{\text{SO}}_{4}$.
Write the answer
The maximum mass of ammonium sulphate that can be produced is calculated as follows:
The maximum amount of ammonium sulphate that can be produced is 2,324 kg.
Calculate the % yield
Example 4: Calculating the mass of reactants and products
Question
Barium chloride and sulphuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.
${\text{BaCl}}_{2}+{\text{H}}_{2}{\text{SO}}_{4}\to {\text{BaSO}}_{4}+2\text{HCl}$
If you have 2 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?
Answer
Find the number of moles of barium chloride
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 0,0096 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
$\text{m}=\text{n}\times \text{M}=0,0096\text{mol}\times 98,12\text{g}\xb7{\text{mol}}^{1}=0,94\text{g}$
(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 ($2\times 0,0096\text{mol}$), which equals $0,0192\text{mol}$.
Find the mass of hydrochloric acid
$\text{m}=\text{n}\times \text{M}=0,0192\text{mol}\times 36,46\text{g}\xb7\text{mol}=0,7\text{g}$
(answer to 2)
Exercise 1: Stoichiometry
Diborane, ${\text{B}}_{2}{\text{H}}_{6}$, was once considered for use as a rocket fuel. The combustion reaction for diborane is:
${\text{B}}_{2}{\text{H}}_{6}\text{(g)}+3{\text{O}}_{2}\text{(g)}\to 2\text{H}\text{B}{\text{O}}_{2}\text{(g)}+2{\text{H}}_{2}\text{O}\left(\ell \right)$
If we react 2,37 g of diborane, how many grams of water would we expect to produce?
The molar mass of diborane is:
$2(10,81)+6(1,01)=27,68\text{g}\cdot {\text{mol}}^{1}$
The number of moles of diborane is:
$n=\frac{m}{M}$ 
$n=\frac{2,37}{27,68}$ 
$n=0,086\text{mol}$ 
For every mole of diborane we get two moles of water and so the number of moles of water is:
$2(0,086)=0,17\text{mol}$
The molar mass of water is:
$15,99+2(1,01)=18,01\text{g}\cdot {\text{mol}}^{1}$
And the mass of water is:
$m=nM$ 
$m=(0,17)(18,01)$ 
$m=3,06\text{g}$ 
Sodium azide is a commonly used compound in airbags. When triggered, it has the following reaction:
$2{\text{NaN}}_{3}\text{(s)}\to 2Na\text{(s)}+3{\text{N}}_{2}\text{(g)}$
If 23,4 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?
The molar mass of sodium azide is:
$23,0+3(14,0)=65,0\text{g}\cdot {\text{mol}}^{1}$
The number of moles of sodium azide produced is:
$n=\frac{m}{M}$ 
$n=\frac{23,4}{64,99}$ 
$n=0,36mols$ 
For every two moles of sodium azide we get three moles of nitrogen gas. This gives the number of moles of nitrogen gas:
$(0,36)\frac{3}{2}=0,54\text{mols}$
At STP one mole of gas would occupy $22,4{\text{dm}}^{3}$ so 0,54 mols occupies: $22,4\times 0,54=12,096{\text{dm}}^{3}$
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 3 mol of carbon dioxide are used up in the photosynthesis reaction, what mass of glucose will be produced?
a) $C{O}_{2}\left(g\right)+{H}_{2}O\left(l\right)light\to {C}_{6}{H}_{12}{O}_{6}+{O}_{2}\left(g\right)$
(note that we put light above the arrow indicating that light is needed to make the reaction take place.)
b) We have 6 carbons on the RHS and only 1 on the LHS so we add a 6 in front of the $C{O}_{2}$:
$6C{O}_{2}+{H}_{2}Olight\to {C}_{6}{H}_{12}{O}_{6}+{O}_{2}$
We also have 12 hydrogens on the RHS and 2 on the LHS so we add a 6 in front of the water:
$6C{O}_{2}+6{H}_{2}Olight\to {C}_{6}{H}_{12}{O}_{6}+{O}_{2}$
Finally we balance the oxygens. There are 18 on the LHS and 8 on the RHS. We add a 6 in front of the oxygen gas and now the equation is balanced:
$6C{O}_{2}\left(g\right)+6{H}_{2}O\left(l\right)light\to {C}_{6}{H}_{12}{O}_{6}+6{O}_{2}\left(g\right)$
c) From the balanced equation we see that for every 6 moles of carbon dioxide used we get one mole of glucose. So from 3 moles of carbon dioxide we get 0,5 moles of glucose ($\frac{3}{6}=0,5$ ).
The molar mass of glucose is:
$6(12,011)+12(1,01)+6(15,99)=180,13g\cdot mo{l}^{1}$
So the mass of glucose produced is:
$m=nM$ 
$m=(180,13)(0,5)$ 
$m=90,06g$ 