Quantitative aspects

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EXPERIMENT 11

Quantitative aspectsAim:To experimentally determine the percentage yield of PbO2 that forms from Pb(NO3)2.

Hypothesis:

Any relevant hypothesis which refers to the % yield expected, e.g. when PbO2 forms Pb(NO3)2 and the reaction runs to completion, there will be a yield of 100%.(The hypothesis does not have to be correct!)

Quantitative aspectsSafety measures:

All lead compounds are very poisonous. Which safety measures do you need to take?

Work in a fume cupboard – do not inhale the fumes.Wear gloves – avoid skin contact.Wear safety goggles.Do not allow lead compounds to go into the drains or rubbish bins – they must be stored in a container for collection by experts in waste-removal.

Quantitative aspectsObservations:The dark brown/black precipitate which forms.The precipitate only starts to form about ±10 minutes after the addition of NaOCℓ.

Experiment 1 Experiment 2 Experiment 3

Mass of filter paper (g)Mass of (filter paper + precipitate) (g)Mass of precipitate PbO2(g)

Quantitative aspectsResults:1. Write a balanced chemical equation for the reaction between lead(II)

nitrate and sodium hydroxide.Pb(NO3)2 + 2NaOH → Pb(OH)2 + 2NaNO3

2. Identify the white precipitate. Explain your answer.White precipitate: Pb(OH)2 it cannot beNaNO3 since all nitrates are soluble in water.

Quantitative aspects3. Calculate the number of mol of lead(II) nitrate in the 10 cm3 that was

used.n

c = V ∴ n = cV = (0,5)(0,001) = 0,000 5 mol Pb(NO3)2

4. Calculate the number of mol of sodium hydroxide in the 10 cm3 that was used.n

c = V ∴ n = cV = (0,5)(0,001) = 0,000 5 mol NaOH

Quantitative aspects5. Which is the limiting reactant in the reaction between the lead(II)

nitrate and the sodium hydroxide?Pb(NO3)2 : NaOH

1 : 2∴ the NaOH is the limiting reactant.

6. Calculate the number of mol of lead(II) hydroxide that can form theoretically.

NaOH : Pb(OH)2

2 : 1∴ 0,000 5 : 0,000 25 ∴ 0,000 25 mol Pb(OH)2 can form.

Quantitative aspects7. Write a balanced chemical equation for the reaction between the

lead(II) hydroxide and sodium hypochlorite (NaOCℓ).Pb(OH)2(s) + NaOCℓ(aq) → PbO2(s) + NaCℓ(aq) + H2O(ℓ)

8. Identify the brown precipitate.Brown precipitate: PbO2

9. Use the theoretical yield of lead(II) hydroxide and calculate the theoretical amount (mol) of lead(IV) oxide that can form.

Quantitative aspectsPb(OH)2 : PbO2

1 : 1∴ 0,000 25 : 0,000 25∴ 0,000 25 mol PbO2 can form.

10. Calculate the theoretical yield of lead(IV) oxide in grams.M(PbO2) = 207 + (16 × 2) = 239 g·mol-1

mand n = M ∴ m = nM

= (0,000 25)(239) = 0,06 g PbO2 form theoretically.

Quantitative aspects11. Calculate the percentage yield of lead(IV) oxide.

true yield% yield = theoretical yield × 100

= 0,06 × 100 = ??,?? %

Conclusions:The percentage yield of lead(IV) oxide was:_______________________________.Give possible explanations for the low yield.There is a loss of product during the process.Not all the reactants are converted to products.The temperature at which the reaction occurs is too low.