Question

3 moles of sodium reacted with 3 moles of chlorine gas. How much material was left unreacted? How many moles of sodium chloride were produced?

Answer 1

Here's what I got.

Explanation:

You need to start by writing the balanced chemical equation that describes the synthesis of sodium chloride

`color(red)(2)"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))`

The balanced chemical equation tells you the mole ratio in which the reactants a re consumed and in which the product is formed.

Notice that the reaction consumes `color(red)(2)` moles of sodium metal for every mole of chlorine gas that takes part in the reaction, and produces `2` moles of sodium chloride.

In your case, the reaction starts with `3` moles of sodium and `3` moles of chlorine gas. Right from the start, you can say that you're going to have insufficient sodium to allow for all the moles of chlorine gas to react.

Why is that the case?

Well, if `3` moles of chlorine gas are to react, then the reaction must consume

`3 color(red)(cancel(color(black)("moles Cl"_2))) * (color(red)(2)color(white)(a)"moles Na")/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "6 moles Na"`

You have less sodium metal than would be needed for all the chlorine gas to react. This means that sodium acts as a limiting reagent, i.e. it is completely consumed before all the moles of chlorine get the chance to react.

So, if all the sodium is consumed, you can say that the reaction also consumed

`3 color(red)(cancel(color(black)("moles Na"))) * "1 mole Cl"_2/(color(red)(2)color(red)(cancel(color(black)("moles Cl"_2)))) = "1.5 moles Cl"_2`

The remaining moles of chlorine gas are in excess, meaning that they are unreacted.

`"moles of Cl"_2color(white)(a)"in excess" = overbrace("3 moles")^(color(blue)("what you start with")) - overbrace("1.5 moles")^(color(purple)("what is consumed"))`

`"moles of Cl"_2color(white)(a)"in excess" = color(green)(|bar(ul(color(white)(a/a)color(black)("1.5 moles Cl"_2)color(white)(a/a)|)))`

Now, the reaction consumes `3` moles of sodium, which means that it produces

`3 color(red)(cancel(color(black)("moles Na"))) * "2 moles NaCl"/(color(red)(2)color(red)(cancel(color(black)("moles Na")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("3 moles NaCl")color(white)(a/a)|)))`