[Frederick95]

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: 

1.  Start with a balanced equation
                                                       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]

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.

[Frederick95]

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]

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.

[Frederick95]

Can you explain with an example please?

[azmanam]

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

[ARGOS++]

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]

yes, yes you are correct.  thank you for the correction.  typing before thinking = bad

[Frederick95]

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]

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

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.

[Frederick95]

Hence the true answer for NaOH is 0.03 mols and not 0.300 mols?

[azmanam]

0.030 mol is what I get.

[Frederick95]

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