Question

Assuming an efficiency of 29.40%, how would you calculate the actual yield of magnesium nitrate formed from 123.5 g of magnesium and excess copper(II) nitrate in the equation `Mg + Cu(NO_3)_2 -> Mg(NO_3)_2 + Cu`?

Answer 1

`"221.5 g Mg"("NO"_ 3)_2`

Explanation:

The first thing to do here is to calculate the theoretical yield of the reaction, i.e. what you'd expect to see for a reaction that has a `100%` yield.

Notice that for your reaction

`"Mg"_ ((s)) + "Cu"("NO"_ 3)_ (2(aq)) -> "Mg"("NO"_ 3)_ (2(aq)) + "Cu"_ ((s))`

every `1` mole of magnesium metal that takes part in the reaction consumes `1` mole of copper(II) nitrate and produces `1` mole of aqueous magnesium nitrate.

Convert the mass of magnesium to moles by using the element's molar mass

`123.5 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305color(red)(cancel(color(black)("g")))) = "5.081 moles Mg"`

Since copper(II) nitrate is in excess, you can assume that all the moles of magnesium take part in the reaction.

This means that at `100%` yield, you would get

`5.081 color(red)(cancel(color(black)("moles Mg"))) * ("1 mole Mg"("NO"_ 3)_ 2)/(1color(red)(cancel(color(black)("mole Mg")))) = "5.081 moles Mg"("NO"_ 3)_2`

Now, you know that the reaction has a `29.40%` yield, which means that for every `100` moles of magnesium nitrate that could be produced, only `29.40` moles are actually produced.

Therefore, this reaction will produce

`5.081 color(red)(cancel(color(black)("moles Mg"("NO"_ 3)_ 2))) * ("29.40 moles Mg"("NO"_ 3)_ 2)/(100color(red)(cancel(color(black)("moles Mg"("NO"_ 3)_ 2))))`

`= "1.4938 moles Mg"("NO"_ 3)_ 2`

To convert this to grams, use the compound's molar mass

`1.4938 color(red)(cancel(color(black)("moles Mg"("NO"_ 3)_ 2))) * "148.3 g"/(1color(red)(cancel(color(black)("mole Mg"("NO"_ 3)_ 2)))) = color(darkgreen)(ul(color(black)("221.5 g")))`

The answer is rounded to four sig figs.