# Question e780b

Mar 23, 2016

Here's what I got.

#### Explanation:

Always start by examining the balanced chemical equation

$\textcolor{red}{3} {\text{Mn"_text((s]) + color(blue)(2)"O"_text(2(g]) -> "Mn"_3"O}}_{\textrm{4 \left(s\right]}}$

Solid manganese will react with oxygen gas to form solid manganese(II, III) oxide in a $\textcolor{red}{3} : \textcolor{b l u e}{2}$ mole ratio.

This mole ratio will help you decide which of the two reactants acts as a limiting reagent.

So, the balanced chemical equation tells you that every time this reaction takes place, it consumes $\textcolor{red}{3}$ moles of manganese for every $\textcolor{b l u e}{2}$ moles of oxygen gas that take part in the reaction.

The ratio that exists between the reactants is true regardless of how many moles of each you have mixed together.

Now, let's take part (a). In this case, you are told that the reaction vessel contains $3$ moles of manganese and $3$ moles of oxygen gas.

According to the aforementioned mole ratio, in order for all the moles of manganese to take part in the reaction, you would need to have

3color(red)(cancel(color(black)("moles Mn"))) * (color(blue)(2)color(white)(a)"moles O"_2)/(color(red)(3)color(red)(cancel(color(black)("moles Mn")))) = "2 moles O"_2

As you can see, you have more oxygen than you actually need. This means that oxygen gas is in excess.

In other words, manganese will be the limiting reagent because it doesn't allow for all the oxygen to take part in the reaction, i.e. it's limiting the amount of oxygen that can react.

In order for all the oxygen to react, you would need

3color(red)(cancel(color(black)("moles O"_2))) * (color(red)(3)color(white)(a)"moles Mn")/(color(blue)(2)color(red)(cancel(color(black)("moles O"_2)))) = "4.5 moles Mn"

Since you only have $3$ moles of manganese, it will be completely consumed before all the oxygen gets a chance to react.

Now, the theoretical yield of a chemical reaction represents the amount of product that you should get if all the moles of reactants that take part in the actual reaction are converted to moles of product.

Notice that the balanced chemical equation shows that when $\textcolor{red}{3}$ moles of manganese react with $\textcolor{b l u e}{2}$ moles of oxygen gas, you get $1$ mole of manganese(II, III) oxide.

In your case, you have $3$ moles of manganese reacting with $2$ moles of oxygen gas, so the reaction will theoretically produce $1$ mole of manganese(II, III) oxide.

Therefore, the theoretical yield for part (a) is

"theoretical yield" = color(green)(|bar(ul(color(white)(a/a)"1 mole Mn"_3"O"_4color(white)(a/a)|)))#

I will leave the other two points to you as practice.

Remember, use the mole ratio that exists between the reactants to see if you're dealing with a limiting reagent, then use the mole ratio that exists between one of the reactants and the product to find the theoretical yield of the reaction.

Hint for parts (b) and (c) $\to$ manganese is the limiting reagent.