Question

Part I: 1. The reaction in Part 1 takes place as shown below. Balance the reaction and de- termine the mass (theoretical yield) of copper(II) nitrate that should be able to be recovered from your copper sample. Show your calculations. 4 HNO,(aq) + | Cu(s) → 1 Cu(NO),(aq) + 2 H,0(0) + Z NO.(Q)

Given an initial 0.128 g Cu (II) and 3mL of HNO3

Answer 1

We have to balance the following equation:

HNO3 + Cu→ Cu(NO3)2 + H2O + NO2

(The states of the reactant and product have not been written since they are irrelevant for balancing).

While balancing, we have to try to make the number of atoms/ions of each element same on both sides. Generally, O, H and N are balanced last since they tend to occur in multiple reactants and products.

Balancing using hit and trial method we get:

4HNO3 + Cu→ Cu(NO3)2 + 2H2O + 2NO2

To solve the next part, the following points must be kept in mind:

· To calculate molar mass of a compound, add the atomic masses of each component atom.

· Equ1: No of moles =

Mass of substance(ing) Molar mass inmor

· To find how much substance is produced or consumed, always compare the number of moles from the balanced chemical reaction.

· Theoretical yield is the amount of product obtained when one of the reactants (limiting reagent) has been completely exhausted.

Balanced reaction :

4HNO3 + Cu→ Cu(NO3)2 + 2H2O + 2NO2

0.128g of Cu is consumed. Molar mass of Cu = 63.5 g/mol

Using the above formula, no. of moles of Cu = = 2.016 x 10-3 moles

0.1289 63.5g/mol

From the balanced equation, 1 mole Cu gives 1 mole Cu(NO3)2

Thus, 2.016 x 10-3 moles Cu gives = 2.016 x 10-3 moles Cu(NO3)2.

** 2.016 x 10-3

Mass of Cu(NO3)2(molar mass = 187.5g/mol) (using equ1) = 2.016 x 10-3 moles x 187.5g/mol = 0.378g

So, theoretical yield = 0.378 g