How much product? Mass in a reaction

How much will I get?

A baker has a recipe: $2$ cups of flour and $1$ cup of sugar make $3$ cakes. If the baker has $10$ cups of flour, the baker can work out how much sugar is needed, and how many cakes will come out. A balanced chemical equation is a recipe in the same way. It tells us how much of each substance is used and made.

Counting with an equation

A balanced equation is more than a list of substances. The coefficients give the ratio of moles. For

$2$ moles of magnesium react with $1$ mole of oxygen to make $2$ moles of magnesium oxide. With the molar masses, we can change this to masses.

• $2\,\text{Mg}$: $2 \times 24 = 48\ \text{g}$
• $\text{O}_2$: $32\ \text{g}$
• $2\,\text{MgO}$: $2 \times 40 = 80\ \text{g}$

The mass on the left, $48 + 32 = 80\ \text{g}$, equals the mass on the right. This is the law of conservation of mass again.

Notice that the numbers of moles are in the ratio $2 : 1 : 2$, but the masses are in the ratio $48 : 32 : 80$. The moles give the ratio, and the masses come from the molar masses. A common mistake is to use the coefficients as masses. They are not masses.

Steps for a mass problem

1. Write the balanced equation.
2. Change the given mass to moles: $n = \dfrac{m}{M}$.
3. Use the ratio of the coefficients to find the moles of the substance that you want.
4. Change these moles to a mass: $m = n \times M$.

Example 1. What mass of magnesium oxide is made when $18\ \text{g}$ of magnesium burns?

• Moles of Mg: $\dfrac{18}{24} = 0.75\ \text{mol}$.
• The ratio of Mg to MgO is $2 : 2$, so the moles of MgO are $0.75\ \text{mol}$.
• Mass of MgO: $0.75 \times 40 = 30\ \text{g}$.

Example 2. Heating $50\ \text{g}$ of calcium carbonate: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$.

• Moles of $\text{CaCO}_3$: $\dfrac{50}{100} = 0.5\ \text{mol}$.
• The ratio is $1 : 1 : 1$, so $0.5\ \text{mol}$ of CaO and $0.5\ \text{mol}$ of $\text{CO}_2$ are made.
• Mass of CaO: $0.5 \times 56 = 28\ \text{g}$. Mass of $\text{CO}_2$: $0.5 \times 44 = 22\ \text{g}$.
• Check: $28 + 22 = 50\ \text{g}$, the same as the mass that we started with.

Example 3. Zinc and sulphuric acid: $\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2$. From $32.5\ \text{g}$ of zinc: $\dfrac{32.5}{65} = 0.5\ \text{mol}$ of $\text{Zn}$ gives $0.5\ \text{mol}$ of $\text{H}_2$, which is $0.5 \times 2 = 1\ \text{g}$ of hydrogen.

Example 4. A ratio that is not 1 : 1. Hydrogen and oxygen make water: $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$. How much water is made from $8\ \text{g}$ of oxygen?

• Moles of $\text{O}_2$: $\dfrac{8}{32} = 0.25\ \text{mol}$.
• The ratio of $\text{O}_2$ to $\text{H}_2\text{O}$ is $1 : 2$, so the moles of water are $2 \times 0.25 = 0.5\ \text{mol}$.
• Mass of water: $0.5 \times 18 = 9\ \text{g}$.

When one reactant runs out

In a real reaction, you may not have the reactants in exactly the right ratio. One of them is used up first, and the reaction stops. That reactant is the limiting reactant. The other is in excess, and some of it is left over.

Example. $2\,\text{Mg} + \text{O}_2 \rightarrow 2\,\text{MgO}$. A student burns $6\ \text{g}$ of magnesium ($0.25\ \text{mol}$) in $4.8\ \text{g}$ of oxygen ($0.15\ \text{mol}$).

• $0.25\ \text{mol}$ of Mg needs only $0.125\ \text{mol}$ of $\text{O}_2$.
• So magnesium is the limiting reactant, and oxygen is in excess by $0.025\ \text{mol}$, which is $0.8\ \text{g}$.
• The product is $0.25\ \text{mol}$ of MgO, which is $10\ \text{g}$, and not more.

The limiting reactant decides the amount of the product.

You should find that adding more of the reactant that is left over makes no more magnesium oxide.

Practise with new numbers

Try it in Python

This program finds the masses in the burning of magnesium.

Common mistakes

• Using the coefficients as masses. The coefficients give moles. Convert to masses with the molar masses.
• Using an unbalanced equation. The ratio will be wrong.
• Forgetting to divide by the molar mass first. Go from grams to moles to moles to grams.
• Forgetting the limiting reactant. If two masses are given, check which one runs out first.

Remember these

• The coefficients of a balanced equation give the ratio of moles.
• Steps: mass to moles, then the ratio, then moles to mass.
• The total mass of the reactants equals the total mass of the products.
• The limiting reactant is the one that is used up first. It decides how much product is made.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What do the coefficients of a balanced equation tell us: masses or moles?

2. In $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$, what mass of oxygen reacts with $36\ \text{g}$ of magnesium?

3. What is a limiting reactant?

4. $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$. What mass of carbon dioxide forms from $6\ \text{g}$ of carbon?