Balanced Equations, Stoichiometry and Percentage Yield Presentation

Balanced Equations, Stoichiometry and Percentage Yield

Learning Goal and Success Criteria

We are learning how to calculate percentage yield. Why: As this is a comparison calculation to determine how reliable experimental data can be. Success Criteria: To be able to . Determine limiting reactants. . Discriminate between experimental and theoretical yield. . Analyse data to determine percentage and theoretical yield. (Formula: percentage yield (%) = experimental yield × 100 theoretical yielsDescribe and explain

Definitions and Calculations

1. Define:

   a molar ratio b limiting reactant c excess reactant.

2. Consider the reaction:

   Ba3N2 + 6H20 -> 3Ba(OH)2 + 2NH3. Calculate the mass of NH3 produced if 220 g of barium nitride (Ba3N2) reacts with excess water.

3. If 120 g of propane (C3Hg) is burnt in excess oxygen,

   calculate the mass of water produced. C3Hg + 502 -> 3CO2 + 4H20

Application, Analysis and Interpretation

1. 90.0 g of iron chloride (FeCl2) reacts with 52 g of

   hydrogen sulfide (H2S) to produce iron sulfide and hydrochloric acid. a Write a balanced chemical equation for the reaction. b Determine the limiting reactant. c Calculate the mass of iron sulfide produced in the reaction.

2. Rescue workers use self-contained breathing

   apparatus to remove carbon dioxide and water from their exhaled breath. First, potassium oxide removes water from the exhaled breath: KO2 + H20 -> O2 + KOH The potassium hydroxide (KOH) produced is then available to react with the carbon dioxide: KOH + CO2 -> KHCO3 a Deduce how much potassium oxide is needed to produce 235 g of oxygen. b If 123 g of KO2 is available in the breathing apparatus, calculate what mass of carbon dioxide can be removed from the exhaled air.

FIGURE 4 Rescue workers use breathing apparatus

Investigation, Evaluation and Communication

1. The chemical equation for photosynthesis is

   6CO2 + 6H20 -> C6H1206 + 6O2. Synthesise the concepts covered in this chapter to explain why plants in a desert fail to produce enough glucose (CRH12OR) to grow.

Q4. FeCl3 + 1.5 H S Calculation

90g 52g Moles FeCl3 = 90 g = 0.555 162.2 amu 52 g Moles HOS = = 1.5257 34.082 amu Using FeCl3 0.555 x 1.5n = 0.83 H2S needed 1n Therefore limiting reagent = FeCl3 as we have 1.5257 moles which is more than 0.83 moles. FeS + 3HCl g = ? Ratio FeCl3 : FeS = 1: 1 Therefore moles FeS expected =moles of FeCl3 = 0.555 grams of FeS = moles x mol mass = 0.555 x 87.91 = 48.79 g

Q5a K2O + H2O -> O2 + KOH Calculation

Grams = ? 235 grams

Q5a K2O + H2O -> 2KOH Calculation

Grams = ? 123 grams KOH + CO -> KHCO3 Grams = ? Grams = ?

Percentage Yield

Percentage Yield Definition and Formula

• The balanced equation (ratios of reactants to products) gives us

  the theoretical yield based on the original mass used.

• Actual yield is the mass produced in the reaction.

  obtained in experiment actual yield × 100 Percent yield = theoretical yield calculated

Example: Phosphorous Trichloride Synthesis

A chemist set up a synthesis of phosphorous trichloride, PCl3, by mixing 12.0g of Phosphorous with 35.0g of chlorine gas and obtained 42.4 g of PCl3. The equation for the reaction is: 2P(s) + 3Cl2 (g) -2PCl3 (l) Calculate the percentage yield of this compound.

Steps for Percentage Yield Calculation

1. Calculate initial moles of reactants
2. Determine limiting reagent
3. Use limiting reagent to determine theoretical moles produced.
4. Use theoretical moles to calculate theoretical mass
5. Calculate percentage yield.

Causes of Error in Percentage Yield

Causes of Percentage Yield Less Than 100%

• Material sticking to glassware.
• Evaporation of a volatile (low

  boiling point compared to room temperature) compound.

• Loss in process of precipitation

  and filtration.

• Occurrence of a competing

  reaction or reactions that compete with the desired reaction.

• Occurrence of a reversible

  reaction

Causes of Percentage Yield More Than 100%

• Contamination of another

  chemical

• Inaccurate weighing of

  equipment and chemical.

Practice: Check Your Learning 8.5

Describe and Explain

1. Explain the difference between theoretical yield

   and experimental yield.

2. Explain why a percentage yield of 109% is impossible.

Apply, Analyse and Interpret

1. Magnesium oxide is formed when magnesium ribbon

   burns in the presence of oxygen: 2Mg + 02 -+2MgO a Determine the theoretical yield of MgO formed when 12 g of Mg reacts with excess oxygen. b When a student performed this experiment, they found 18 g of MgO was formed. Determine the percentage yield of this reaction.

2. Heptane undergoes a combustion reaction with

   oxygen to form carbon dioxide and water: C,H16 + 1102 -> 7CO2 + 8H20 a Determine the theoretical yield of CO2 formed when 10 g of C,H16 is burnt with excess O2. b A chemist analysing this reaction produced 25 g of carbon dioxide. Determine the percentage yield for this reaction.

Investigate, Evaluate and Communicate

1. When a wine bottle has a defective seal, the

   ethanol (C2H,OH) becomes exposed to oxygen. As a result, acetic acid (CH,COOH) and water are formed. a Construct a balanced chemical equation for this reaction. b Decide if ethanol or oxygen would be the limiting reactant in this reaction. Justify your response. c Investigate methods used by wine makers to prevent this reaction from occurring. Present your findings to your peers.