Question

Need some help with a Chemistry question!

Answer 1

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

Answer 2

2mole of NH3 is 34.08grams

Answer 3

So, isn't it 2 moles of ammonia?

Answer 4

Yes, NH3 is ammonia.

Answer 5

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

Answer 6

You know limiting reactant?

Answer 7

Yes, the limiting reactant is Nitrogen.

Answer 8

No, I'm sorry it's hydrogen.

Answer 9

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

Answer 10

It can produce 102.24 grams of ammonia.

Answer 11

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

Answer 12

Dont need grams.
My formula is a shorter way.

Answer 13

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

Answer 14

Okay, so it can produce 3.3 moles of NH3

Answer 15

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

Answer 16

Yea.

Answer 17

Makes sense, since 3mol H2 produces 2 mol NH3

Answer 18

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

Answer 19

5 mol H2 would produce around 3.3 mol NH3

Answer 20

No problem!

Answer 21

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

Answer 22

@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.

Answer 23

Okay, thank you!! That makes sense.

Answer 24

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.

Answer 25

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.

Answer 26

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

Answer 27

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

Answer 28

is this clearer?

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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.

Answer 35

yes B is the limiting reagent

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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.

Answer 41

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