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Stoichiometry

In our daily lives, from the food we eat to the fuel that powers our vehicles, stoichiometry plays a crucial role in ensuring that chemical processes are efficient and effective. Improve your knowledge of stoichiometry by using this resource to learn about the stoichiometric coefficient, the molar ratio and chemical equations.

Stoichiometry describes the quantitative relationship between relative quantities of reactants and products in a balanced chemical equation. Stoichiometric coefficients are the numbers written immediately before the atoms, ions and molecules in a balanced chemical equation. These numbers indicate the ratio of any reactant or product that reacts or forms in a reaction.

As an example, let's consider the reaction between hydrogen gas and nitrogen gas to form ammonia. The balanced chemical equation can be written as:

3H2(g)+N2(g)2NH3(g)

The numbers three, one and two are written in front of H, N and NH3, respectively. These are the stoichiometric coefficients. They show that three molecules of hydrogen gas react with one molecule of nitrogen gas to form two ammonia molecules. When there is no coefficient before an atom, ion or molecule, we interpret it as one. For example, the coefficient of molecular nitrogen in this reaction is one.

Molar ratios

The stoichiometric coefficients in a balanced chemical equation can also be interpreted on the molar scale. Molar scale interpretation relates to the quantities of substances involved in the reaction. Therefore, if we know the amount of one substance involved in the reaction, we can use the molar ratio to determine the amounts of others involved in the reaction. Let's consider the same example:
3H2(g)+N2(g)2NH3(g)

Using the molar scale, we can say that three moles of hydrogen gas react with one mole of nitrogen gas to form two moles of ammonia gas.

Limiting and excess reactants

In chemical reactions, limiting and excess reactants (or reagents) determine the extent of the reaction and the amount of products formed. The limiting reactant is the substance that is consumed completely first, thus limiting the amount of product that can be formed. Once the limiting reactant is used up, the reaction cannot proceed further, even if other reactants are still available. On the other hand, excess reactants are those that remain after the reaction has reached completion.

Let's consider the reaction between hydrogen gas and nitrogen gas to form ammonia gas again:
3H2(g)+N2(g)2NH3(g)

Based on stoichiometry, we know that if we mix three moles of H2 with one mole of N2, we can obtain two moles of NH3. In other words, H2 reacts with N2 in a molar ratio of 3:1.

If we mix three moles of hydrogen gas and two moles of nitrogen gas, two moles of NH3 will be produced. However, the entire quantity of H2 will be used before the N2 as we start with the exact amount of H2 required for the reaction. The product mixture will contain one unreacted mole of N2 as we only required one mole of N2 for the reaction. If we had more H2, we could have produced more NH3 as we already have excess N2. Therefore, in this reaction, we call H2 the limiting reagent since it limits the production of NH3.

Hence, the limiting reagent is the reactant that runs out first, limiting the formation of products. Nitrogen gas is called the excess reagent because we have a greater amount than required for the reaction.

Reaction yield

The theoretical yield is the maximum amount of product that can form under the given reaction conditions. This is determined by the amount of limiting reagent available. Theoretical yield can be calculated using the stoichiometric coefficients (or molar ratios) in the balanced chemical equation. We calculate theoretical yield assuming no losses occur during the reaction, and all the reactants are converted into products. However, in reality, losses and inefficiencies occur during chemical reactions. Therefore, the actual amount you isolate and measure from a chemical reaction is less than what you expect in theory.

The actual yield of a reaction is the amount of the product you isolate from the chemical reaction. Actual yield must be determined experimentally, so it often referred to as experimental yield. You need to isolate the product and weigh it to get the actual yield.

The percentage yield indicates how efficient the reaction was in converting reactants into products. If the value is closer to 100%, most of the reactants were converted into products; the reaction was closer to completion. Percentage yield can be calculated using:

Percentage yield=Actual yieldTheoretical yield×100%

Example 1 – calculating the amount of product formed from a chemical reaction

Calculate the amount in moles of ammonia gas that can be produced from 0.18 mol of hydrogen gas. 3H2(g)+N2(g)2NH3(g)

Step 1: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the "unknown") is written as the numerator at the top, and the coefficient for the compound we have the amount for (the "known") is written as the denominator at the bottom.

Stoichiometric ratio=coefficient of unknowncoefficient of known=23

Step 2: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.

n(NHA3)=23×0.18 mol=0.12 mol

The burning of methane gas (CH4) in excess oxygen produces carbon dioxide gas and water vapour. The balanced chemical equation for the reaction is:
CH4(g)+2O2(g)CO2(g)+2H2O(g)

Calculate the amount of water vapour that can be obtained from the complete burning of 3.20 g of methane gas. The molar mass of methane is 16.0 g mol1.

Step 1: Calculate the amount in moles of methane burned.

n(CHA4)=mM=3.20 g16.0 g mol1=0.200 mol

Step 2: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the "unknown") is written as the numerator at the top, and the coefficient for the compound we have the amount for (the "known") is written as the denominator at the bottom.

Stoichiometric ratio=coefficient of unknowncoefficient of known=21

Step 3: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.

n(HA2O)=21×0.200 mol=0.400 mol

Example 3 – calculating the theoretical yield of a reaction

Phosphoric acid is produced by treating phosphate rock, which contains calcium phosphate, with sulfuric acid:
Ca3(PO4)2+3H2SO43CaSO4+2H3PO4
A chemist in a fertiliser production plant mixes 144.5 g of calcium phosphate with 160.5 g of sulfuric acid. Determine the maximum mass of phosphoric acid that can be formed. The molar masses are given in the table.

Compound Molar mass (g mol1)
Ca3(PO4)2 310.2
H2SO4 98.1
H3PO4 97.99

Step 1: Identify the limiting reagent in this reaction by first, finding the number of moles of each reactant.

n(Ca3(PO4)2)=mM=144.5 g310.2 g mol1=0.4658 mol

n(H2SO4)=mM=160.5 g98.1 g mol1=1.64 mol

Step 2: Compare the calculated amounts of two reactants. The easiest way to find the limiting reactant is to use the formula:

ncoefficient

The reactant with the smallest ncoefficient is the limiting reactant.

For Ca3(PO4)2:

ncoefficient=0.4659 mol1=0.4659 mol

For H2SO4:

ncoefficient=1.64 mol2=0.82 mol

As the ncoefficient for Ca3(PO4)2 is smaller, it is the limiting reactant.

Step 3: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the "unknown") is written as the numerator at the top, and the coefficient for the limiting reactant (the "known") is written as the denominator at the bottom.

Stoichiometric ratio=coefficient of unknowncoefficient of known=21

Step 4: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.

n(HA3POA4)=21×0.4659 mol=0.9317 mol

Step 5: Calculate the mass of product formed. This is the theoretical yield.

m(HA3POA4)=n×M=0.9317 mol×97.99 g mol1=91.29 g

Example 4 – calculating percentage yield

When 64.5 g of methane is burnt in the presence of excess oxygen, it forms 136.2 g of carbon dioxide and water vapour.
CH4(g)+2O2(g)CO2(g)+2H2O(g)

However, the theoretical yield of carbon dioxide for this reaction is 176.4 g. Calculate the percentage yield of carbon dioxide for this reaction.

Step 1: Recall the formula for calculating percentage yield.

Percentage yield=Actual yieldTheoretical yield×100%

Step 2: Substitute the values into the formula to calculate percentage yield.

Percentage yield=136.2 g176.4 g×100%=77.21%

Fe and OA2 react according to the following equation:
4Fe+3O22Fe2O3
When 168.3 g of Fe reacts with O2, 172.6 g of Fe2O3 is obtained. Determine the percentage yield of Fe2O3 for this reaction. The molar mass of Fe2O3 is 159.69 g mol1 and Fe is 55.85g mol1.

Step 1: Calculate the amount in moles of Fe reacted.

n(Fe)=mM=168.3 g55.85 g mol1=3.013 mol

Step 2: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the "unknown") is written as the numerator at the top, and the coefficient for the limiting reactant (the "known") is written as the denominator at the bottom.

Stoichiometric ratio=coefficient of unknowncoefficient of known=24

Step 3: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.

n(FeA2OA3)=24×3.013 mol=0.1507 mol

Step 4: Calculate the mass of product formed. This is the theoretical yield.

m(FeA2OA3)=n×M=0.1507 mol×159.69 g mol1=240.6 g

Step 5: Recall the formula for calculating percentage yield.

Percentage yield=Actual yieldTheoretical yield×100%

Step 6: Substitute the values into the formula to calculate percentage yield.

Percentage yield=172.6 g240.6 g×100%=71.74%


Further resources

Molar mass and stoichiometry

If you need more support working with molar masses and stoichiometry, check out this helpful resource!

Limiting reagents and yield

Explore limiting reagents and yield using this resource.