Question

Question about finding the quantity of the excessive reactant...

Answer 1

For example, let's say the problem was \[C_3H_8+5O_2\rightarrow3CO_2+4H_2O\]
In which, 1.5 moles of C3H8 and 15 moles of O2 are bought together.

Answer 2

So now, I get that you get the ratios of the problem and equation to know which is the reactant and excessive.

problem=0.10
equation=0.2

therefore C3H8 is the limiting reactant.

Answer 3

Now to get the quantity of O2, your suppose to multiply the limiting reactant moles from the PROBLEM (not equation) , so it would be 1.5

Answer 4

But I'm not sure what to multiply 1.5 by..

Answer 5

@abb0t

Answer 6

@Abhisar

Answer 7

@Jhannybean

Answer 8

@Jhannybean

Answer 9

you're given \(\sf 1.5 ~mol~ C_3H_8\) so use that to find the number of moles of water. That will tell you how many moles of water you have through the reactant, \(\sf C_3H_8\).

Do the same with \(\sf 15 ~mol~ O_2\). use that to find the number of molesof water. This will also tell you how many moles of water you have by reacting the oxygen, \(\sf O_2\).

Answer 10

Compare the two and see which one produces less.

Answer 11

Between the two, pick one and find the moles of the other reactant. This will tell you how many moles of it you NEED. compare what you HAVE by what you NEED and you will find your Limiting Reactant.

Answer 12

Not sure..are you telling me to find the limiting reactant? If so, I already know it's C3H8

Answer 13

can we be serious here

Answer 14

continue @Jhannybean

Answer 15

So if you've found what your LR is, use the quantity of the excess reactant given in the problem to find how many moles of the non limiting reactant there are left,

Answer 16

So if \(\sf C_3H_8\) is your LR, then use the \(\sf15~mol~O_2\) to find how many moles of Oxygen did not react.

Answer 17

But your problem should give you how much of the sample was used.

Answer 18

You would use sample amount (g) - amount reacted by the excess (g)

Answer 19

This would give you how much excess reactant remains.

Answer 20

My book teaches like,

1.calculate the mole ratio of the problem and equation
2.determine which reactant is excess and LR
3. calculate the quantity of the excess reactant that reacts with the lr

My ratios are

(problem)=1.5 mol C3H8/15 mol O2=0.10 mol C3H8/mol O2
(equation)= 1/5=0.20 mol C3H8/mol O2

Since the problem is < than the equation, that means that the denominator is excess. Which would be O2.

I'm looking for a pattern and it appears that the quantity used for the excess is

the LR in the mole ratio for the problem (1.5) times the ratio for the equation where the excess is the numerator

so 1.5 x 5/1=7.5 mol O2

is that correct?

Answer 21

You can name it whatever you want. i.e : 1 and 2, have and need, problem and equation...etc

Answer 22

Is is 7.5 mol of O2 and is my statement of the excess should be the numerator correct?

Answer 23

Your question is not making sense to me.

Answer 24

(problem)=1.5 mol C3H8/15 mol O2=0.10 mol C3H8/mol O2
(equation)= 1/5=0.20 mol C3H8/mol O2

Those are my ratios.

According to the patterns in my book. It looks like to find the quantity of the excess used (O2)

You would take the limiting reactant value from the (problem) which is 1.5, and multiply it by the ratio of the equation, (looks like the excess value must be the numerator)

Answer 25

So the equation ratio was originally 1/5, but since the 5 is the excess value, I switched it to 5/1.

1.5 x 5=7.5 O2 mol

Answer 26

am i the only one who is not getting what you are doing?

Answer 27

Im a bit confused too.

Answer 28

The way I learned it isthe way I explained it. Did that make any sense to you?

Answer 29

no
i learnt in a different way too
you make sense while he does not
what is that problem thingy? what is that equation thingy?

Answer 30

ratio of C3H8 to O2 is 1:5
mole of C3H8 = 1.5
1.5 mole - 1
X mole - 5
so X mole = 1.5 * 5 = 7.5 mole of Oxygen are needed to completely react with C3H8
total amount given of O2 is 15 mole
15- 7.5 = EXCESS OXYGEN REACTANT

Answer 31

I'm getting the same answer as you so I guess it's fine. I believe I got it now.

Answer 32

Awesome.

There are innumerous methods of solving limiting reactant problems.

Answer 33

yeah there are
i just find it easy using mathematical way 1.5 - 1
X - 5
since its less time consuming + good for test conditions

Answer 34

btw excess is denominator of the ratio
but since you are dividing 1.5 by 1/5 ratio
according to mathematical rules the fraction will flip upside down and the equation will change from division to multiplication
\[\frac{ 1.5 }{ 1 } \div \frac{ 1 }{ 5 } = \frac{ 1.5 }{ 1 }\times \frac{ 5 }{ 1 }\]

Answer 35

@Somy stop that. You're not a chemist.

Answer 36

@abb0t i dont have to be a chemist for this <.<