Question

How many grams of Cu will form if 15.5g Al react with 85.3 g CuCl₂ according to the following reaction?

2Al(s) + 3CuCl₂(aq) → 2AlCl₃+ 3Cu(s)

Round your answer to one decimal place, even if that doesn't follow significant figure rules. Do not include units.

Examples of the correct format: 13.2 or 5.4

Answer 1

Hi! This is a stoichiometric problem.

This means that the stoichiometry will moderate the amounts involved in the chemical reaction, so let's start by taking a look at it:

2 Al(s) + 3 CuCl₂(aq) → 2 AlCl₃+ 3 Cu(s)

So, we need to find out how many GRAMS of Cu will be formed.

We have information about the amounts of reactives: Al and CuCl₂, but the stoichiometry is necessarily related to moles of molecule or atom, NOT IN UNITS OF MASS, as gram is.

Also there is the concept of limitant reactant: there have to be 2 moles of Al and 3 moles of CuCl₂ for the reaction to happen. If we have less the reaction stops. So we need to find out the ratio between moles of reactants.

For units of moles we need a periodic table in order to know the molar mass of the elements, there are apps that make the addition for you, then just divide the mass by the molar mass and you'll get the moles.

	Al	CuCl2
Molar Mass (g/Mol)	26.982	134.448
Mass (g)	15.5	85.3
moles	0,574	0,634
reactant	in excess	limitant

So, CuCl₂ is the limitant reactant, since: 0,634/3 < 0,574/2; and that is what we'll use to make calculations.

So, 3 moles of CuCl₂ will make 3 moles of Cu. Since the ratio between them is=1 the moles will be the same. Then 0,634 moles of Cu will be produced, and Cu molar mass= 63.546 g/mol.

Then there will be 63.546 g/mol * 0,634mol= 40,3 g

Hope to have been helpful!