- Meehan98

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- Meehan98

1 mole of NH3 is 34.08 gram. Do I just convert that back into moles?

##### 1 Attachment

- anonymous

2mole of NH3 is 34.08grams

- Meehan98

So, isn't it 2 moles of ammonia?

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- anonymous

Yes, NH3 is ammonia.

- Meehan98

How do I figure out how many moles of NH3 are produced in the reaction?

- anonymous

You know limiting reactant?

- Meehan98

Yes, the limiting reactant is Nitrogen.

- Meehan98

No, I'm sorry it's hydrogen.

- anonymous

Then use hydrogen and find the amount of ammonia it can be produced.
Ignore nitrogen then.

- Meehan98

It can produce 102.24 grams of ammonia.

- anonymous

\[\frac{ mol hydrogen }{ 1}\times \frac{ mol ammonia }{ mol hydrogen }\]

- anonymous

Dont need grams.
My formula is a shorter way.

- Meehan98

I did nitrogen again..It would be 56.8 grams of ammonia.

- Meehan98

Okay, so it can produce 3.3 moles of NH3

- anonymous

5 mol H2 x (2mol NH3/3mol H2)

- anonymous

Yea.

- anonymous

Makes sense, since 3mol H2 produces 2 mol NH3

- Meehan98

Thank you! I was making things more difficult than what was needed!

- anonymous

5 mol H2 would produce around 3.3 mol NH3

- anonymous

No problem!

- Meehan98

Quick question: How come we used the limiting reactant (H) instead of the excess reactant (N)?

- Photon336

@Meehan98 you always use the limiting reactant to calculate the number of moles and eventually grams of product formed for one simple reason: the limiting reactant runs out first because there is less of it than your reactant that's in excess. once your limiting reactant runs out you cant form any more product because it runs out.

- Meehan98

Okay, thank you!! That makes sense.

- Photon336

No problem, anytime.. it's just that when one reactant runs out first,
think of it this way
say we have a reaction in which A+B = c
FROM this reaction we need one molecule of A and one of B. to form C.
\[A + B --> C \]
Say if we had 10 molecules of A and only 5 of B.
B would be our limiting reactant because we only have 5 molecules of B and from our reaction we need 1 molecule of A and one molecule of B.
so because of this B runs out first, we use the number of moles of B to calculate A.

- Photon336

sorry for that last part we use the number of moles of B and then multiply by the ratio of C/B to get how much C is produced.

- Meehan98

Okay, what if it's asking how many grams will be produced instead of moles?

- Photon336

So from what we said, you need to identify which reactant runs out first and that's going to be our limiting reactant.
then you multiply the number of moles of the limiting reactant by the molar ratio to get your product in moles
then you can easily say since we have moles of product we multiply that by the molar mass like this:
\[moles*\frac{ (molar mass) g }{ moles }\] = grams of product

- Photon336

is this clearer?

- Meehan98

Yes, it is. Thank you! I'm just having trouble figuring out which is the limited reactant in an equation.

- Photon336

here is a problem for you completely made up though
A molar mass = 10g/mol
B molar mass = 5g/mol
20 grams of A react with 10 grams of B to form C.
A + 2B --> C
1. find me the limiting reagent
2. and how many grams of C are produced.

- Photon336

FYI this is an important step
We have 20 grams of A
and 10 grams of B.
We must first convert everything to moles.
20grams of A * (1mol/10g) = 2 moles of A
10 grams of B*(1mol/5g) = 2 moles of B

- Photon336

So this tells us that we have 2 moles of A and two moles of B.
we must find out how many moles we NEED for each.
to do this we multiply each by the molar ratio.
we look at the formula: we have 1 mole of A and 2 moles of B in he equation.
so for A we do
\[2 A = \frac{ 2B }{ A } = 4 B\]
when we do B
\[2 B * \frac{ A }{ 2B } = 1 mol of A\]
do you notice something? the way I set this

- Photon336

We need 4 moles of B and 1 mol of A but how many moles of each do we have?

- Meehan98

Ok, so I think that I use a much more drawn out way of producing the answer than what is needed, but I got B as the limiting reagent and 10 moles of C.

- Photon336

yes B is the limiting reagent

- Photon336

good, how did you get the number of moles of C?

- Meehan98

But, I still need to convert the moles into grams so that will be 10 moles multiplied by the molar mass (5) = 50?

- Meehan98

I got the number of moles of C by multiplying 5 mol B times the ratio of C/B which is 10.

- Photon336

oh lol sorry in my problem i didn't give you the molar mass of C the imaginary compound; let's say the molar mass of C is 5 grams per mole. show me like by writing out the equation of how you get the grams of C

- Meehan98

Based on your explanation above of how to convert from moles to grams of the product, \[\frac{ 5 mol B }{ 1 } X \frac{ mol C }{2 mol B }\] so, I have 10 mol C and you said to multiply that by the molar mass, does that mean by molar mass of C? If so, then I get 50, but I'm still in moles. I'm sorry- not quite grasping this yet.

- Photon336

|dw:1441916098745:dw|
well first we have identified the limiting reagent which is B.
now we multiply by the molar ratio to find the number of moles of C

- Photon336

Does this make sense?

- Meehan98

Yes, I was using 5 mol of B when I multiplied the ratio instead of 2.

- Photon336

okay so put your information together like this
LIMITING REAGENT B
2 moles of B
Molar ratio of B/C
C/2B why did I put B on the bottom?

- Meehan98

I'm not sure, why?

- Meehan98

Is it because it's the limiting reagent? because I would've written it like: 2molB/1molC

- Photon336

B was our limiting reagent

- Photon336

and we had 2 moles of B

- Photon336

Next we must figure out the number of moles of C

- Photon336

\[2 mol B * \frac{ C }{ 2B }\] = 1 mol C

- Photon336

I want you to look at tis last sentence. when we multiplied by the molar ratio we were looking for C, right? so when we set it up we make sure that B is in the denominator and you can see that B cancels out leaving us with C? make sense?

- Meehan98

Yes, I know how to do that; I just didn't understand now that we have 1 mol C, how to convert to grams like the question asked.

- Photon336

\[1 mol * (\frac{ 5 grams }{ mol } ) = \]

- Photon336

set it up like this general rule
\[moles*(\frac{ molar mass grams }{ mol })\]

- Meehan98

So, C would be 5 grams.

- Photon336

exactly

- Meehan98

Okay, a lot to take in, but helped immensely!! Thank you for spending time and explaining everything out! I truly appreciate it!

- Photon336

no problem

- Photon336

look up dimensional analysis; it will help with these problems

- Meehan98

Ok, will do:)

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