[Tyler Durden]

Ok, so i've been trying to find a way to solve this problem, i feel like im really close to cracking it but i end up with a wrong answer:

"A Student reacted the following unkknown metal, M, with HCl and collected the following data: mass of metal = 0.300g, volume of water displaced= 227.3mL, temperature of H₂ (hydrogen gas) = 25C, Barometric pressure = 760.0torr, volume of HCl= 20.0mL, molarity of HCl = 1.500 mol/L, and volume of NaOH = 36.0mL. If the metal, M, had an atmoic weight of 50.0 and reacted to produce MCl₃, what was the molarity of the NaOH solution?"

Answer is : 0.333 molarity NaOH

Any help would be appreciated. This might help you guys:
-remember of the formula PV=nRT to find any info. on hydrogen gas
-Unbalanced equation is M + HCl------>MCl₃ +H₂
-Pressure of water is 23.8torr so the pressure of H2 is 736.2
-Gas constant, R = 62.4 L*torr/K*mol

[macman104]

I'm confused where does the NaOH come into play?

Was this part of a lab, or was this the actual question (in it's entirety)?

[Tyler Durden]

Its a question from my lab class, its like a practice question more or less on the lab, so it doesnt come directly from the experiment itself. NaOH comes in when you titrate it with HCl to find the excess HCl, after the metal has reacted with it. In the experiment, we balance the chemical equation by finding out the moles of each substance.
Hopefully this helps. The experiment is called "Stoichiometry of the reaction of Magnesium with hydrochloric acid."

[Borek]

Answer is : 0.333 molarity NaOH

That's correct result.

Show your work.

[Tyler Durden]

I cant show my work because i dont know how to get there lol. When i get it i will show my work.
Do you know how to get 0.333 Molarity NaOH?

[macman104]

I cant show my work because i dont know how to get there lol. When i get it i will show my work.

Ok, so i've been trying to find a way to solve this problem, i feel like im really close to cracking it but i end up with a wrong answer

I'm confused did you attempt the problem or not?

[Tyler Durden]

Of course i attempted the problem, only like 20 times. Im still trying to get it. I dont want to show what i've done because i might be doing it all wrong.

From my calculations so far i get:
9.00x10^-3 mol H₂
1.08 g HCl
6.00x10^-3 mol M

I think what i need to do next is balance the chemical equation from the information i have, im not sure, maybe not. Still working on it.

[macman104]

Right, but even if you might be doing it wrong, it shows us what information you do understand and then we can focus on what you don't know.

Start from the beginning.

Write the balanced equation for the reaction of the metal with the acid.

Then, find the number of moles of HCl you added (you have the molarity and volume) and the moles of metal (you have the weight added and the atomic mass).

EDIT:  Also, telling us the answers you get from your calculations is not as useful as showing us your calculations because if the answer you get is wrong, it's hard to tell why you got it wrong.

[Tyler Durden]

Im using unit factors to get the answers, i dont know how to make it look any more neat than this:

Pressure of H₂ = 760.0 torr - 23.8 torr H₂0 =

PV=nRt (Ideal gas Equation)
n=PV/RT
_mol H₂=  (736.2torr)(0.2273L)/(62.4L*torr/K*mol)(298K) = 9.00x10^-3 moles H₂

_mol HCl = 20.0mL HCl * x10^-3L/1mL * 1.500mol HCl/1L = 0.0300 mol H₂

_mol M = 9 0.300g M * 1mol M/50g M= 6.00x10^-3


Chemical equation:

__ M + __ HCl  __ MCl₃ + __ H₂          i havnt balanced it yet

mol ratio of HCl = mol HCl/mol M
mol H₂ = mol H₂/Mol M

Again, im doing something wrong because i dont get the right answer.

The last step is finding the molarity of NaOH, not sure how though.

[Tyler Durden]

Ok, it seems im doing the right steps. Except my mole ratios in the balanced equation is incorrect. The mole ratio of HCl is supposed to be 3 according to my calculations but i keep getting 5. Remember that to find the mole ratio of HCl, its the moles of HCl over the moles of M. If i manage to get this then that would solve this problem.

[Borek]

__ M + __ HCl  __ MCl₃ + __ H₂          i havnt balanced it yet

That's the basis for the whole question. Assuming you start with one atom of M - how many molecules of HCl do you need? How many molecules of H₂ will be produced? Don't be afraid of fractional coefficient at this stage.

The last step is finding the molarity of NaOH, not sure how though.

From the definition - number of moles/volume of the solution.

[Tyler Durden]

I finally got the right answer!!
Turns out i needed some rest lol.

Heres what i did (i will omit the actual problem solving, for it is time consuming and i do not have much time):

Step 1: Found the moles of H₂ in the system, by using the Idea Gas Equation, PV=nRT, where n is moles of H₂
           which would give 9.00x10^-3 moles H₂

Step 2: Found the initial grams of HCl in the reaction, which it is found using the data they give in the problem (20.0mL and 1.500 molarity)
           This should come out to be 1.08g HCl

Step 3: Find the moles of unknown metal M, which is found by using the data given in the problem (.300g of M and atomic wt. 50.0g)
           Should come out to be 6.00x10^-3 mol M

Step 4: Find the number of moles of HCl that reacted with unknown metal M. I found this by using the mol of H₂ from the chemical     
           equation. Since HCl reacts with M and decomposes to H₂, i find how much HCl was used to produce 9.00x10^-3
           H₂. This gives the amount of HCl which reacted in the chemical equation. Its 0.648g HCl

Step 5: Find the excess HCl, which is found by subtracting the initial amount of HCl with the amount which reacted. Step 4 - Step 2
           Should give 0.432g HCl excess

Step 6: Find the molarity of NaOH. Since NaOH is being titrated with the excess, i use the formula, eq_a=eq_b or NxL=g/ew
           Where N is normaility, ew= equivalent weight, L is Liters, g grams
           This gives .333 normality, which when converted to molarity also gives 0.333 Molarity NaOH the correct answer

Chemical Equation should be

1 M + 3 HCl  1.50 H₂ + 1 MCl₃

[Borek]

9.00x10^-3 moles H₂

Correct, no idea what for you have calculated it 

1.08g HCl

Correct, but not necessary. All you need is number of moles.

6.00x10^-3 mol M

Correct and necessary.

0.648g HCl

Why mass? Why not moles?

Should give 0.432g HCl excess

Again - why mass? Why not moles?

0.333 Molarity NaOH the correct answer

Correct.

You start with moles, you end with moles, for some unknown reason you use mass for intermediate steps. You can, but it requires using molar masses and make it easier to make a mistake.

[Tyler Durden]

9.00x10^-3 moles H₂

Correct, no idea what for you have calculated it 

1.08g HCl

Correct, but not necessary. All you need is number of moles.]

I calculated moles of H₂ to find the mole ratios in the chemical equation, mol H₂ over mol M, thats where i got the coefficient of 1.50 in the chemical equation.

I converted HCl to grams because i used to formula N_NaOHxL_NaOH=g_HCl/EW_HCl
There are other methods for it i know, but i use the one i have on my mind at the moment lol.

[Borek]

You know that MCl₃ is produced, that's enough to balance the equation:

2M + 6HCl = 2MCl₃ + 3H₂

You started with 0.300g of M, with molar mass of 50.0g - that means 0.300/50.0 moles.

From the reaction stoichiometry it must have reacted with 3*0.300/50.0 moles of HCl.

You started with 20.0mL of 1.500 mol/L HCl, or 20.0/1000*1.500 moles of HCl.

That means there was 20.0/1000*1.500 - 3*0.300/50.0 excess moles of HCl.

To neutralize it you needed the same number of moles of NaOH.

You used 36.0/1000 liters of NaOH solution. Its concentration was (20.0/1000*1.500 - 3*0.300/50.0)/(36.0/1000).

That's the shortest and the safest way to get to the final result.