Stoichiometric Calculations
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 section relate to each other and to the balanced chemical equation:
Example: Stoichiometric Calculation
Question
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))
Step 1: Write the balanced equation
\[\text{C}_{3}\text{H}_{8}\text{(g)} + 5\text{O}_{2}\text{(g)} \rightarrow 3\text{CO}_{2}\text{(g)} + 4\text{H}_{2}\text{O (g)}\]Step 2: 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 \(5:1\).
Step 3: 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.
Example: Stoichiometric Calculation
Question
What mass of iron (II) sulphide is formed when \(\text{5.6}\) \(\text{g}\) of iron is completely reacted with sulphur?
Step 1: Write the balanced equation
\[\text{Fe (s)} + \text{S (s)} \rightarrow \text{FeS (s)}\]Step 2: 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*}Step 3: 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.
Step 4: 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 \(\text{8.79}\) \(\text{g}\).
This lesson is part of:
Quantitative Aspects of Chemical Change