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A student performs the following lead extraction to test the stoichiometric method and to determine the efficiency of the reaction. In the reaction 49.5 g of lead(II) nitrate is reacted with excess zinc in a single displacement reaction.

a. Calculate the theoretical yield of lead in the experiment.
b. After separating the lead by filtration and allowing it to dry, the student obtains 19.8g of lead. What is the percent yield of the experiment?

Sagot :

Theoretical yield produced is 31.8 gram and the percentage yield is 62.27%.

The zinc (Zn) is more reactive than lead (Pb). So that zinc loses electrons more easily than lead .

And the electrons of Zn are transfer to lead in this reaction .

This question is a limiting reactant question .

And For limiting reactant questions so that we need to determine that how much product can be produced by all reactant .

Zinc and Lead (II) nitrate react to form Zinc Nitrate and Lead.

Zn + Pb(NO₃)₂ → Zn(NO₃)₂ + Pb

Now we have- Weight of Pb(NO₃)₂ = 49.5 g

molar mass of Pb(NO₃)₂ = 331.2 g

now moles of Pb(NO₃)₂ = 49.5 ÷ 331.2 = 0.15 moles

Moles of Pb consumed = 1 × 0.15 = 0.15

moles mass of Pb consumed = 0.15 × 207.2 = 31.8 gm

Here Zn is in excess so that lead(II) nitrate is a limiting reactant.

(a) Theoretical yield of lead -  

moles of Pb produced = 0.15 × 1  = 0.15 moles

molar mass of Pb = 207.2g

mass of Pb produced in reaction = 0.15 × 207.2   = 31.08 g

theoretical yield = 31.8 gram

(b) After separating the lead by filtration -

obtained lead = 19.8gm

Percentage yield = 19.8 ÷ 31.8 ×100 = 62.27%

so that after separating the lead the percentage yield is 62.27%.

So, After applying the concepts of stoichiometric method the theoretical yield produced came out to be 31.8 gram and the percentage yield came out to be 62.27%.

Learn more about theoretical yield here:

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