Chapter 5: Stoichiometry Part 2 SABIS Grade 11 (Level M) Chemistry

 

• Chapter 5: Stoichiometry

  
  5.2 Excess and Yield

5.2.1 Excess and Limiting Reagents

In reactions that go to completion, if we are given only one quantity of a reactant or product, all other quantities of reactants consumed and products produced by the reaction can be calculated. If two or more quantities of reactants are given, most probably they will not be in the exact ratio in which they should react. In this case, the one found in the smaller relative quantity will react completely, and it is called the limiting reagent. The other one will not react completely and is called the reagent in excess. The following is such an example:

Example 1

1.50g of H2 gas is mixed with 6.0 dm3 of O2 at RTP and the mixture is exploded. How much of which gas is left, how many moles of water will be produced and how much energy is evolved? (Enthalpy of formation of gaseous water is ΔH = – 242 kJ/mole; 1 mole of gas at RTP occupies 24 dm3).

Solution

First, summarize the given and required to find:

Given: 1.50 g H2; 6.0 dm3 O2 at RTP.

RTF: (a) Which element reacts completely? (b) How much is left of the other?

 (c) How many moles of water are produced? (d) How much heat is evolved?

Solution: In general,

            aA          +          bB       →        cC

     a moles of A + b moles of B → c moles of C

A and B react in the mole ratio of (aA/bB). If the given mole ratio of A to B is larger than the reaction ratio aA/bB , then A will be in excess. Otherwise, A will react completely.

           2H2(g)         +          O2(g)          →    2H2O(g)               ΔH = – 484 kJ

    2 moles of H2(g) + 1 mole of O2(g)   →    2 moles of H2O(g)

(a =2, b = 1)

{Given mole ratio} -???- {Reaction mole ratio}      Which is larger?

The given mole ratio = (1.5g×1mol/2g)H2(6.0dm3×1mol/24dm3)O2${\displaystyle \frac{(1.5\mathrm{g}\times 1\mathrm{m}\mathrm{o}\mathrm{l}/2\mathrm{g}){\mathrm{H}}_2}{(6.0\mathrm{d}{\mathrm{m}}^3\times 1\mathrm{m}\mathrm{o}\mathrm{l}/24\mathrm{d}{\mathrm{m}}^3){\mathrm{O}}_2}}$ = 0.75 / 0.25 = 3.

The reaction mole ratio is a/b = 2/1 = 2

(a) Which element reacts completely?

Since the given H2(g) to O2(g) ratio is more than the reaction ratio, H2(g) is in excess and O2(g) is the limiting reagent which reacts completely.

(b) How much is left of the other?

We use oxygen to determine how much hydrogen reacts and how much water is produced. The given quantity x of O2(g), 6.0 dm3, or 0.25 mol, will react completely.

Quantity of H2(g) used up = 0.25 mol O2(g) × {2 mol H2(g) / 1 mol O2(g)}

   = 0.25 × (2/1) = 0.50 mol H2(g)

Quantity of H2(g) left = 0.75 – 0.50 = 0.25 mol.

(c) How much H2O(g) is produced?

Quantity of H2O(g) produced = 0.25 mol O2(g × {2 mol H2O(g) / 1 mol O2(g)}

= 0.25 × (2/1) moles = 0.50 mol H2O(g)

(d) How much heat is produced?

Quantity of heat produced = 0.50 mol H2O(g) × {242 kJ / 1 mol H2O(g)} = 1.2 × 102 kJ

5.2.2 Percentage Yield

The maximum amount of product that can be produced in a reaction is called the theoretical yield. The actual quantity produced is called the actual yield. The percentage yield is defined as:

Percentage yield = actualyieldtheoretical yield$\frac{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}$ × 100

The percentage yield is always less than 100% because (1) not all reactions go to completion and (2) frequently unwanted reactions called “side” reactions take place consuming some of the reactants.

Example

H2 gas reacts with carbon monoxide, CO, to produce methanol CH3OH. When 4.00 kg of H2 gas reacts completely with CO gas, the amount of methanol produced was 0.75 × 103 mol. Calculate the percentage yield.

Solution

First, summarize the given and required to find:

Given: 4.00 kg H2 produced 0.75 × 103 mole CH3OH.

RTF: Percentage yield

Solution: Percentage yield = actual yieldtheoretical yield${\displaystyle \frac{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}}$ × 100

The actual yield is given. To find the percentage yield we must calculate the theoretical yield, which is the maximum amount of CH3OH that can be produced.

Step 1: Write the balanced equation:

                   2H2(g)        +           CO(g)      →          CH3OH(g)

          2 moles of H2(g) + 1 mole of CO(g) → 1 mole of CH3OH(g)

Step 2: Find the reaction ratio:

The reaction ratio we should use is (1 moles CH3OH) / (2 moles of H2), or one mole to two moles.

Step 3: Find the theoretical yield:

Theoretical quantity of methanol produced =

(4.0 × 103g) / (2.0 g per mole) of H2 × (1mol CH3OH / 2mol H2) = 1.0 × 103 mole

Step 4: Find the percentage yield:

Percentage yield = actual yieldtheoretical yield${\displaystyle \frac{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}}$ × 100 = (0.75 × 103 mole) / (1.0 × 103 mole) = 75 %

H2 gas reacts with carbon monoxide, CO, to produce methanol, CH3OH. When 0.50 tons of H2 gas reacts completely with CO gas, the amount of methanol produced was 4.0 tons. Calculate the percentage yield.

Given Molar mass in g.mol−1: H = 1, C = 12, O = 16

Write your answer to three significant figures.

Percentage yield is ......  % ?

70 kg of N2 and 40 kg of H2 were allowed to react. Calculate the maximum mass of NH3 that may be produced. If in the above reaction 34 kg of NH3 were produced, calculate the percentage yield of the reaction.

Given Molar mass in g.mol−1: H = 1, N = 14

Write your answer to two significant figures.

Percentage yield is....... % ?

Back to Chapter 5 Menu: Stoichiometry

Move to Chapter 5 Part 3 The Manufacture of Sulfuric Acid