## Excess Reactant Help [ENDORSED]

Posts: 113
Joined: Wed Sep 30, 2020 9:51 pm

### Excess Reactant Help

How do you find how much excess reactant will be left over?

Kelly Yun 2I
Posts: 103
Joined: Wed Sep 30, 2020 10:03 pm

### Re: Excess Reactant Help  [ENDORSED]

Hi,

You would first figure out the moles of the limiting reactant and then balance the equation to find out the mole ratio of each reactant in the chemical reaction. Then, assuming that all of the limiting reactant is used up, we would be able to find out how much of the reactant in excess would be consumed during the equation. Sometimes, the question may just say that the reactant is in excess (without giving a specific amount) in which we may not be able to find out the amount excess reactant left. However, if they give us the amount of the excess reactant (i.e. 1.00 kg), then we can just subtract the amount consumed in the reaction from the large amount to see how much will be left after the reaction.

Tessa House 3A
Posts: 91
Joined: Wed Sep 30, 2020 9:52 pm
Been upvoted: 1 time

### Re: Excess Reactant Help

To start, find out which reactant is the limiting reactant. Once you know this, convert grams or moles of limiting reactant to grams of the excess reactant by using the mole ratio of the reactants found in the balanced chemical equation and the molar mass of the excess reactant. Do this to find out how many grams of the excess reactant reacted with the limiting reactant to find out how many grams were consumed in the reaction. Then, subtract this number from the number of grams of excess reactant you were given in the problem to find out how many grams of excess reactant remain.

Pranav Daggubati 3C
Posts: 109
Joined: Wed Sep 30, 2020 9:35 pm

### Re: Excess Reactant Help

To start a problem to find excess reactant, remember that it's all about mole ratios. Let's say you are given the equation: $2\textup{Al} + 3\textup{Cl}_2 \rightarrow \textup{2AlCl}_3$ and it says you react 255 g of $\textup{Al}$ with 535 g of $\textup{Cl}_2$ to get an actual yield of 300 g of product. First, you take the 255g and convert it to moles, which is 9.45 mol of $\textup{Al}$. Do the same for the 535 g $\textup{Cl}_2$ to get 7.54 mol.

Now, because you know that you need $3\textup{Cl}_2$ for every $2\textup{Al}$, its easy to see that the $\textup{Al}$ is in excess. This is because you have $9.45$ mol, which is much more than $\frac{2}{3}$ of $7.54$ mol of $\textup{Cl}_2$. Therefore the $\textup{Al}$ is in excess.

To find how much excess you have, you have to do some simple arithmetic. First, multiply $9.45 \cdot \frac{2}{3}$ and subtract that from $9.45$. Now you have the moles of excess $\textup{Al}$, which you can now use to convert to whatever measure of quantity you would like to.

Hope this made sense and hope it helps!