Question

Question #52b36

Answer 1

The theoretical yield of product is 2.17 g ( w/ sig figs).

Explanation:

First we need to determine the limiting reactant.

Answer 2

Here's what I got.

Explanation:

I'm not really sure I understand what you did there, so I'll just break the problem down step by step and you can use the solution as a guide.

The idea here is that you need to figure out if one of the two reactants will act as a limiting reagent.

You know that aluminium and chlorine gas react according to the balanced chemical equation

`color(blue)(2)"Al"_ ((s)) + color(purple)(3)"Cl"_ (2(g)) -> 2"AlCl"_ (3(s))`

The reaction consumes `color(purple)(3)` moles of chlorine gas for every `color(blue)(2)` moles of aluminium metal that take part in the reaction.

So, convert the masses of the two reactants to moles by using their respect molar masses

`0.439 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(27color(red)(cancel(color(black)("g")))) = "0.01626 moles Al"`

`2.29 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(71color(red)(cancel(color(black)("g")))) = "0.03225 moles Cl"_2`

Now, pick one of the reactants. Let's pick aluminium. Use the `color(blue)(2) : color(purple)(3)` mole ratio that exists between aluminium and chlorine gas to see if you have enough moles of chlorine gas available.

`0.01626 color(red)(cancel(color(black)("moles Al"))) * (color(purple)(3)color(white)(.)"moles Cl"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles Al")))) = "0.02439 moles Cl"_2`

This is the number of moles of chlorine gas needed to ensure that all the moles of aluminium react. As you can see, you have more than enough moles of chlorine gas available

`overbrace("0.03225 moles Cl"_2)^(color(brown)("what you have"))" " > " " overbrace("0.02439 moles Cl"_2)^(color(brown)("what you need"))`

You can thus say that chlorine gas is in excess, which implies that aluminium will act as a limiting reagent, i.e. it will be completely consumed before all the moles of chlorine gas can react.

Now, to find the theoretical yield of the product, simply use the number of moles of the limiting reagent and the mole ratio that exists between the limiting reagent and the product.

In your case, aluminium is the limiting reagent, so you'll have

`0.01626 color(red)(cancel(color(black)("moles Al"))) * "2 moles AlCl"_3/(color(blue)(2)color(red)(cancel(color(black)("moles Al")))) = "0.01626 moles AlCl"_3`

To convert this to grams, use the molar mass of the product

`0.01626 color(red)(cancel(color(black)("moles AlCl"_3))) * "133.5 g"/(1color(red)(cancel(color(black)("mole AlCl"_3)))) = color(darkgreen)(ul(color(black)("2.17 g")))`

The answer is rounded to three sig figs.