Chemistry » Quantitative Aspects of Chemical Change » Stoichiometric Calculations

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


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


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 \(\text{8.79}\) \(\text{g}\).

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