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#?

1 Answer
Mar 3, 2017

Answer:

#"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.