Question

Mg + 2H2O --> Mg(OH)2 + H2

If you mix 5.0 grams of Mg and 5.0 grams of H2O

This is the balanced equation, I need to find the limiting agent, theoretical yield, and percent yield. Thanks

Answer 1

So far, I have the moles for Mg and H2O, Mg = .206 moles, H2O = .277 moles.

Answer 2

For one mole of Mg, how many moles of H2O do you need?

Answer 3

Ik the ratio is 2 H2O - 1Mg, but would it be .277*2?

Answer 4

That is the correct ratio. We first need to find out the limiting reagent.

The limiting reagent is the reactant that gets all used up in the reaction. The amount of product that can form is limited due to the limiting reagent because while there may be excess other products, our limiting reagent has been used up.

Answer 5

If we have .206 moles of Mg, how many moles of \(\text{H}_2\text{O}\) are needed to completely react with magnesium?

Remember the 1:2 ratio

Answer 6

I think H2O is the limiting because .412 is needed but we only have .277 available. I used the 2-1 ratio to find how much H2O we would need (.412)

Answer 7

Bingo!

Answer 8

Yup thank you! now I just need the theoretical yield part, which is what I'm having trouble with the most

Answer 9

Now to calculate the theoretical yield (I'm assuming of \(\text{Mg(OH)}_2\))

our reaction is
\(\text{Mg + 2H}_2\text{O} \to \text{Mg(OH)}_2 + \text{H}_2\)

If we're using 0.277 moles of \(\text{H}_2\text{O}\), how many moles of Mg will we be using up?

Answer 10

I'll explain the x and 2x in a minute once you tell me how many moles of Mg will react with 0.277 moles of \(\text{H}_2\text{O}\)

Answer 11

Is it the original .206 moles?

Answer 12

Nope. If use 0.206 moles of Mg, then we'll need 0.416 moles of \(\text{H}_2\text{O}\)

And we only have 0.277 moles!
Since it's a 1:2 relation, to find out how many moles of Mg will react with water, we just have to divide by 2.

What is 0.277/2 = ?

Answer 13

OOHHHHHH that makes sense so it's .1385

Answer 14

I could have also done .416 - .277 right?

Answer 15

No, coincidentally it comes out to be the same number but 0.416 is what we calculated as the amount of \(\text{H}_2\text{O}\) needed to react IF we were trying to react all of the 0.206 moles of magnesium.

Answer 16

Since water is our limiting reagent, we know we're going to use up all 0.277 moles of that. The next step involves figuring out how much of the 0.206 moles of magnesium will react- which you calculated to be 0.1385 moles

Answer 17

So we just divide the ratio from the moles of the limiting reactant. for example if the ration was 1:1 we would have just done .277/1 I'm assuming

Answer 18

Yes! Because that would be the mole to mole ratio :)

Answer 19

That makes a lot of sense

Answer 20

So like I said, we're using up all the water which is 0.277 moles
But for magnesium, we're using half the number of moles so we're using up 0.1385 moles out of the 0.206 moles of magnesium.

Answer 21

I just added the numbers but I left the last two columns for you. How many moles of the product would we yield?

Answer 22

.0675 moles?

Answer 23

We had 0.206 moles and then we used up 0.1385 moles of magnesium. That's how we got 0.0675 moles.

For our products, we started off with 0. And we produce 0.1385 moles.

0 + 0.1385 = ?

Answer 24

.1385

Answer 25

and that's how many moles of our product you form!

We can convert that to grams but it doesn't specify in your question if they're fine with it being in moles.

Answer 26

Did they give you a value for the actual yield?

Answer 27

Thank you! yes they did but I think I understand how to find the percent yield pretty well. It's just actual/theoretical * 100. I do have some questions still on how you found the theoretical yield if that's fine

Answer 28

Absolutely, ask away!

Answer 29

How did you get .1385 for both the products? and for the theoretical yield, would it be for the precipitate?

Answer 30

So remember that everything in the chemical reaction is moles to moles

We react 1 mole of Mg with 2 moles of \(\text{H}_2\text{O}\) to form 1 mole of \(\text{Mg(OH)}_2\) and 1 mole of \(\text{H}_2\)

We found that \(\text{H}_2\text{O}\) is our limiting reagent which means it gets all used up.

Since we need 2 moles of that to form one mole of our products, that means
0.277 moles / 2 = 0.1385 moles

We use up 0.1385 moles of Magnesium with 0.277 moles of water to form 0.1385 moles of our products.

It's still in that 1:2:1:1 mole ratio

Does that clarify it better?

Answer 31

To answer your second question, yes. That's how many moles is formed for both the precipitate and hydrogen gas \((\text{H}_2)\)

But you would be calculating the percent yield of the precipitate

Answer 32

Yea that clarifies it a bit, but for my first question, why would the products both be .1385? So we used up all of the .277 from H2O but what happened to the .1385 from Mg?

Answer 33

Because .277/2 = .1385 which accounts for both products, and that leaves the .1385 from Mg

Answer 34

The 0.1385 moles of Mg react with the 0.277 moles of \(\text{H}_2\text{O}\)

Answer 35

The remaining 0.206-0.1385 = 0.0675 moles of Magnesium will not have reacted because there was no more water left to react with

Answer 36

Ohhhhhh, Ic. I was looking at it the wrong way. It's the .0675 that is left unreacted because H2O is used up already. The .277 accounts for the .1385 in both the products.

Answer 37

And lastly I would have to turn the moles .1385 into grams because the actual yield is in grams. Thanks for your help!

Answer 38

Yup! And usually in real life, you have a whole bunch of different methods to purify the precipitate and not have any other contaminants but that's a whole different world out there.

Answer 39

That's correct! You have to find the molar mass of \(\text{Mg(OH)}_2\) and convert it to grams :)

Answer 40

Thanks again! I understand this a lot better

Answer 41

It was my pleasure!