IMAT 2014 Q44 [Percentage Yield]

Excess lead (II) nitrate solution is added to 1.30 g of zinc powder and the mixture is stirred. When the reaction is finished the lead formed is filtered, dried and weighed. It has a mass of 3.31g.

What is the percentage yield of the lead?

[A_r: Pb = 207; Zn = 65]

A. 60%
B. 70%
C. 100%
D. 90%
E. 80%

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The first step is to construct a chemically balanced equation to represent this reaction.

We should know that the formula for Lead (||) Nitrate is Pb(NO_3)_2 and Zinc Nitrate is Zn(NO_3)_2 , respective to their valence’s.

Zn + Pb(NO_3)_2 \rightarrow Pb +Zn(NO_3)_2

Next, we should try to understand the moles reacting and moles produced. The most important thing to note here, would be molar ratios. From the formula we can assume that if 1 mole of zinc reacts, 1 mole of lead is produced.

This makes calculations easier, since we have the amount of grams of Zinc reacting.

If one mole of zinc is 65g, then x moles is 1.30g. Using basic algebra we know that x is 0.02 moles.

And so, theoretically, 0.02 moles of lead should also be produced.

Our last steps are to gather what the actual yield of lead produced is. If one mole of Lead is 207g then x moles is 3.31g. X is 0.016 moles of Lead that were actually produced.

This leads to the percentage yield of Lead, which should be calculating using:

\frac {Actual Yield}{Theoretical Yield} \times 100%

\frac {0.016}{0.02} \times 100% = 80%

Making our answer, E.