Answer to Question #138583 in General Chemistry for Jaylene

Question

Assuming an efficiency of 28.40%, calculate the actual yield of magnesium nitrate formed from 128.9 g of magnesium and excess copper (II) nitrate

Answer 1

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(No3)2(aq)$\to$ Mg(No3)2(aq)+Cu(s)

Every 1 mole magnesium that takes part in the reaction consumrs cuper(2) nitrate and produce 1mole of aquous magnesium nitrate

Convert the mass of magnesium to moles by using the elements moler mass

(128.9g)(1 mole of Mg)/24.305g

=5.30 mole of Mg

Since Cupper(2)nitrate is in excess,you can assume that all the moles of magnesium takes place in reaction

That means that at 100% yield, you would get. (5.30 moles of Mg)(1 mole of Mg(No3)2/(1 moles of Mg)

=5.30 moles of Mg(No3)2

Now the reaction has 28.40% yield. Which means that for every 100 moles magnesium nitrate that could be produced, only 28.40 moles are actually produced.

So,this reaction will produce

( 5.30 moles Mg(No3)2)(28.40 moles Mg(No3)2)/100 moles Mg(No3)2

=1.505 moles Mg(No3)2

we will covert the mass use of moler mass of Mg(No3)2=148.3 g/mol

=(1.505)(148.3) g of Mg(No3)2

=223.22 g of Mg(No3)2