September 28, 2023, 12:24:39 AM
Forum Rules: Read This Before Posting

### Topic: Determine the limiting reagent  (Read 11672 times)

0 Members and 1 Guest are viewing this topic.

#### Frederick95

• Regular Member
• Posts: 82
• Mole Snacks: +0/-12
##### Determine the limiting reagent
« on: March 07, 2008, 06:04:12 PM »
Example 4:
Copper (II) sulfate (50.0 mL/0.300 M) react with sodium hydroxide (50.0 mL/0.600 M) in a double displacement reaction.  The initial temperature of each solution is 21.4oC.  A precipitate of copper (II) hydroxide and aqueous sodium sulfate are produced.  Determine the enthalpy change of this reaction if the highest temperature reached is 24.6oC.

Solution:

CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)

2.  Determine the volume of water
Since the volume of each solution is 50.0 mL, the volume of water is assumed to be 100.0 mL and thus its mass is 100.0 g.

3.  Calculate the heat transferred
q = mcDT
= (100 g)(4.18 J/goC)(3.2oC)
= -1.34 x 103 J

Notice that the value for q is a negative number.  Since the temperature increased this means the reaction was exothermic.

4.  Determine the limiting reagent
# mols CuSO4 = 0.0150 mols; # mols NaOH = 0.300 mols
Since these ratios are 1:2, as they are in the balanced equation, there is no limiting reagent.

MY QUESTION IS HOW DOES THIS ANSWER GET 0.015 mols AND 0.300 mols?

THANK YOU

#### azmanam

• Chemist
• Sr. Member
• Posts: 1417
• Mole Snacks: +160/-24
• Mediocrity is a handrail -Charles Louis d'Secondat
##### Re: Determine the limiting reagent
« Reply #1 on: March 07, 2008, 06:51:32 PM »
Let's take a look at the information you were given.

CuSO4: 0.300 molar solution, 50 milliliters.
NaOH: 0.600 molar solution, 50 milliliters.

We aren't given any other information about the initial amounts of reagents, so we'll have to make that work.

There is a formula for determining molarity, do you know what it is?  In fact, we know 2 of the 3 variables in that formula, and can solve for the third.  This equation will get us well on our way to answers of 0.015 and 0.300 mols, respectively.

hint: watch your units, and make sure you have the right units when you do your calculations.
Knowing why you got a question wrong is better than knowing that you got a question right.

#### Frederick95

• Regular Member
• Posts: 82
• Mole Snacks: +0/-12
##### Re: Determine the limiting reagent
« Reply #2 on: March 07, 2008, 07:40:09 PM »
That makes no sense since:

molarity = moles of solute / litres of solution

= 0.600 / 0.05 L

= 12

There number of mols of NaOH would be 12, which it is not (its 0.3 M)

#### azmanam

• Chemist
• Sr. Member
• Posts: 1417
• Mole Snacks: +160/-24
• Mediocrity is a handrail -Charles Louis d'Secondat
##### Re: Determine the limiting reagent
« Reply #3 on: March 07, 2008, 08:11:28 PM »
Ah.  0.600 is not the moles of solute.  It is the molar concentration.

You did a good job remembering that the volume should be 0.05 L, though.  that is what everyone misses.
Knowing why you got a question wrong is better than knowing that you got a question right.

#### Frederick95

• Regular Member
• Posts: 82
• Mole Snacks: +0/-12
##### Re: Determine the limiting reagent
« Reply #4 on: March 07, 2008, 08:22:11 PM »
Can you explain with an example please?

#### azmanam

• Chemist
• Sr. Member
• Posts: 1417
• Mole Snacks: +160/-24
• Mediocrity is a handrail -Charles Louis d'Secondat
##### Re: Determine the limiting reagent
« Reply #5 on: March 07, 2008, 09:27:41 PM »
Molarity is moles of solute per liter of solvent (not solution), as you indicated.

M = mol/L

When you have any 2 of 3, you can solve for the third.

mol = M*L
L = mol/M

So which 2 do we have here?  Which are we solving for?  Limiting reagent problems are based on number of moles.  Sometimes you're given grams.  Sometimes you're given volume (if it's a gas), sometimes density or molarity if it's a liquid or a solution.

Granted that, we are looking for number of moles.  We have a solution in this problem, which is why you have correctly indicated we need the molarity formula.

One other piece of information that may be useful is the units of molarity.  The units of molarity are mol/L.  Thus, one indication your first guess may not have been correct is your units - using the formula you provided, (0.600 mol/L)/(0.05L) = (12 mol/L2)  The units of your answers are a good indication of whether you're on the right track or not.

Can you use this information to give it another try?j
Knowing why you got a question wrong is better than knowing that you got a question right.

#### ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
##### Re: Determine the limiting reagent
« Reply #6 on: March 07, 2008, 09:35:11 PM »

Dear Azmanam;

Sorry!, -  You did a little mistake.
Molarity is moles of solute per liter of solvent (not solution), as you indicated.

M = mol/L
MolaRity =  Number of Moles per Liter Solution.

Good Luck!
ARGOS++

#### azmanam

• Chemist
• Sr. Member
• Posts: 1417
• Mole Snacks: +160/-24
• Mediocrity is a handrail -Charles Louis d'Secondat
##### Re: Determine the limiting reagent
« Reply #7 on: March 07, 2008, 09:37:48 PM »
yes, yes you are correct.  thank you for the correction.  typing before thinking = bad
Knowing why you got a question wrong is better than knowing that you got a question right.

#### Frederick95

• Regular Member
• Posts: 82
• Mole Snacks: +0/-12
##### Re: Determine the limiting reagent
« Reply #8 on: March 07, 2008, 09:41:01 PM »
mol = L x M
=0.05 L x .300 mol
=0.0150 mols

Therefore, the number of mols is CuSO4 is 0.0150 mols

mol = L x M
= 0.05 L x .600 mol
=0.0300 mols

But why does the answer say the number of moles of NaOH is 0.300 moles?

#### azmanam

• Chemist
• Sr. Member
• Posts: 1417
• Mole Snacks: +160/-24
• Mediocrity is a handrail -Charles Louis d'Secondat
##### Re: Determine the limiting reagent
« Reply #9 on: March 07, 2008, 09:47:06 PM »
The numbers in your calculations are correct, well done.

The units I have a bit of a problem with.  M stands for molarity, and units of molarity are mol/L.  you indicate M as just mol.  Otherwise, it looks good

Quote
Since these ratios are 1:2, as they are in the balanced equation, there is no limiting reagent.

If this is a quote from an answer key, then the numbers in the answer key should be 0.015 and 0.030.  0.30 is a wrong answer (perhaps a typo?).  But if you're quoting the 'no limiting reagent' line, then your calculations are correct.
Knowing why you got a question wrong is better than knowing that you got a question right.

#### Frederick95

• Regular Member
• Posts: 82
• Mole Snacks: +0/-12
##### Re: Determine the limiting reagent
« Reply #10 on: March 07, 2008, 09:48:58 PM »
Hence the true answer for NaOH is 0.03 mols and not 0.300 mols?

#### azmanam

• Chemist
• Sr. Member
• Posts: 1417
• Mole Snacks: +160/-24
• Mediocrity is a handrail -Charles Louis d'Secondat
##### Re: Determine the limiting reagent
« Reply #11 on: March 07, 2008, 09:52:15 PM »
0.030 mol is what I get.
Knowing why you got a question wrong is better than knowing that you got a question right.

#### Frederick95

• Regular Member
• Posts: 82
• Mole Snacks: +0/-12
##### Re: Determine the limiting reagent
« Reply #12 on: March 07, 2008, 09:56:52 PM »
Thank you so much!

I have one more question that I would really like help with I have been spending about 3 days on this question alone and remain stumped

Use the experimental values to calculate the enthalpy change

Given
Quantity of reactant 1: 5.61 g KOH(s)
Quantity of reactant 2: 200.0 mL of 0.50 M HBr(aq)
Initial Temperature (oC): 20.0
Final Temperature (oC): 26.7

Required
ΔH = ?

My work thus far
ΔT = tf – ti
= 26.7 oC - 20.0 oC
= 6.7 oC