# 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?

Mar 3, 2017

"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.

${\text{Mg"_ ((s)) + "Cu"("NO"_ 3)_ (2(aq)) -> "Mg"("NO"_ 3)_ (2(aq)) + "Cu}}_{\left(s\right)}$

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

$1.4938 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Mg"("NO"_ 3)_ 2))) * "148.3 g"/(1color(red)(cancel(color(black)("mole Mg"("NO"_ 3)_ 2)))) = color(darkgreen)(ul(color(black)("221.5 g}}}}$

The answer is rounded to four sig figs.