2Al + 3F -> 2AlF3 What will be present after the reaction and how much when 40g of aluminium and 57g of fluorine react?

Answer 1

So a helpful term to consider here might be ‘limiting reagent’—the reactant that is totally consumed when a chemical reaction is completed. If this doesn’t ring a bell, it essentially means that when you consider the molar quantities of every reactant in a particular experimental setup, one of them will be used up first, preventing the reaction from proceeding any further (as one of the reactants is now used up/missing).

So what’s the limiting reagent here?

Well, we might think aluminum to start as there are only 40g, but this isn’t necessarily the case. First, the rate of consumption of aluminum and fluorine isn’t the same. Secondly, we’re given the mass of each reactant, not the amount in moles.

So first, let’s convert each reactant into moles.

The molar mass of aluminum is 26.982 g/mol. Given that there are 40g of Al, there are thus 1.48 moles of aluminum.

The molar mass of fluorine is 18.998. (Importantly, I think you meant to write 3F2, not 3F. Fluorine is highly reactive and exists in the F2 gaseous form, not as a solo atom. In addition, the equation isn’t balanced unless you have 3F2.) Given this, the molar mass of diatomic fluorine is 37.997 g/mol. Given that there are 57g of fluorine, there are 1.50 moles of diatomic fluorine.

At first glance, there appears to be the same number of moles of Al and F2. However, we should note that the diatomic fluorine and aluminum are not consumed at the same rate. For every 2 Al molecules consumed, 3 F2 molecules are also consumed. Thus, F2 is our limiting reagent.

So given that the diatomic fluorine is used up, we know that we will have some remaining aluminum. Since diatomic fluorine is consumed at a 3:2 ratio compared to aluminum, 1.5 moles of diatomic fluorine being reacted means 1.0 moles of aluminum are reacted. In that case, we should have 1.48-1.0=0.48 moles of aluminum left.

However, we’re also producing a product when we consume reactants. 3 equivalents of diatomic fluorine (and 2 equivalents of aluminum) yield 2 equivalents of AlF3. Thus, for every 1.5 moles of diatomic fluorine reacted, we produce 1 mole of AlF3.

Depending on your instructor’s preferences, you can leave your answer as is or you may be asked to convert from moles to grams. This last step is straightforward—just multiply by the molar mass.