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Offline Bob Sacamano

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Stoichiometry Problem
« on: September 25, 2009, 11:12:45 AM »
A 0.9157 g mixture of CaBr2 and NaBr is dissolved in water, and AgNO3 is added to the solution to form AgBr precipitate. If the mass of the precipitate is 1.6930 g, what is the percent by mass of NaBr in the original mixture.

I am confused since there are two compounds of Br and it doesn't tell how much of each one there is. I tried to write a long equation with all products and reactants but since it is a combination of two reactions it appears it can be balanced an infinite number of ways.

Is there a different way to approach this problem?

Offline Borek

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Re: Stoichiometry Problem
« Reply #1 on: September 25, 2009, 11:15:27 AM »
Try to use information given to write two equations in two unknowns.
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Offline Bob Sacamano

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Re: Stoichiometry Problem
« Reply #2 on: September 25, 2009, 11:28:08 AM »
I get the two equations

CaBr2 + 2AgNO3 --> 2AgBr + Ca(NO3)2

NaBr + AgNO3 --> AgBr + NaNO3

Now since I know the mass of the precipitate I can find that there are 9.01e-3 mol of AgBr produced. However now I don't know how much AgBr was yielded by the NaBr.

Offline Borek

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Re: Stoichiometry Problem
« Reply #3 on: September 25, 2009, 12:19:45 PM »
You are correct so far, although you have missed the point. Assume your unknowns are masses of CaBr2 and NaBr. Think in terms of algebraic equations now, not a reaction equations. Using these unknowns, can you write equation for the total sample mass? Can you write equation for the mass of AgBr produced?
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Offline Bob Sacamano

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Re: Stoichiometry Problem
« Reply #4 on: September 25, 2009, 12:36:59 PM »
ok... I'll give it a try.

For every 3 mol of AgBr yielded I would think 1 mol of NaBr would be consumed.

(9.01e-3 mol AgBr) (1/3) = 0.3092 g NaBr consumed

0.3092 g / 0.9157 g = 33.77 % NaBr by mass

This is not correct.

Offline Borek

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Re: Stoichiometry Problem
« Reply #5 on: September 25, 2009, 02:30:11 PM »
For every 3 mol of AgBr yielded I would think 1 mol of NaBr would be consumed.

No, for each mole of AgBr one mole of NaBr was consumed - that's from teh stocichiometry of the reaction.

Can you try to follow what I wrote earlier, or will you be trying to guess the correct solution combining given numbers in all possible ways? If the latter, it is not worth my time. Try to express given information in terms of algebraical equations, starting with mNaBr and mCaBr2 - these are your unknowns.
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Offline Bob Sacamano

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Re: Stoichiometry Problem
« Reply #6 on: September 25, 2009, 08:50:12 PM »
I understand what you mean but I don't see how I can formulate the equations for the mass of sodium bromide and calcium bromide without being given the composition of the original solution.

Offline Borek

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Re: Stoichiometry Problem
« Reply #7 on: September 26, 2009, 04:39:40 AM »
Not equations for the masses, but equations that will express given information in terms of masses given. Like

mNaBr + mCaBr2 = 0.9157 g

is a mathematical way of stating that

0.9157 g mixture of CaBr2 and NaBr
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Offline Bob Sacamano

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Re: Stoichiometry Problem
« Reply #8 on: September 26, 2009, 12:17:19 PM »
Ok so now I have

mNaBr = 0.9157 g - mCaBr2

mCaBr2 = ?

But how can I create an equation for the mass of CaBr2 using only the the variable for the mass of sodium bromide and the moles of silver bromide?



Offline Borek

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Re: Stoichiometry Problem
« Reply #9 on: September 26, 2009, 12:50:44 PM »
What is mass of AgBr produced from mNaBr?
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Offline Bob Sacamano

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Re: Stoichiometry Problem
« Reply #10 on: September 26, 2009, 01:54:47 PM »
You have to realize I have worked on this single problem for several hours and have explored every possible way of solving it that I have been taught. I have the answer to the question (56%), that is not what I am looking for if that's why you are so reluctant to give me anything but the steps to solve it. I am pretty sure this question was assigned for practice by accident because my instructor specifically mentioned that we would not be doing to two variable questions. That being said, I am more than willing to invest time in learning how to solve this type problem and I'm sorry I don't understand what you mean but you have to understand that I simply do not have the knowledge to work it out.

Offline Borek

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Re: Stoichiometry Problem
« Reply #11 on: September 26, 2009, 02:56:25 PM »
What would be mass of AgBr produced form 1 g of NaBr?
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Offline Bob Sacamano

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Re: Stoichiometry Problem
« Reply #12 on: September 26, 2009, 03:46:11 PM »
(1 g NaBr) * (mol NaBr / 102.9 g) * (mol AgBr / mol NaBr) * (187.8 g / mol AgBr) = 1.83 g

Offline Borek

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Re: Stoichiometry Problem
« Reply #13 on: September 26, 2009, 05:22:43 PM »
Can't you use exactly the same approach to express mass of AgBr in terms of mNaBr? In terms of mCaBr2?

You will get something like

mAgBr = (some numerical constant)*mNaBr
« Last Edit: September 27, 2009, 02:58:23 PM by Borek »
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Offline Bob Sacamano

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Re: Stoichiometry Problem
« Reply #14 on: September 27, 2009, 12:16:32 PM »
I got this answer from someone else but it is off by 4% I can't see why it isn't correct?

This is a tough one, but I am going to try to explain it as best as I can (hope your algebra is good!).

First thing to realize here is that there are 2 equations we have to deal with:
CaBr2 + 2AgNO3 = Ca(NO3)2 + 2AgBr
NaBr + AgNO3 = NaBr + AgBr

For the first equation:
(0.9157-x)g of CaBr2 produces (1.693-y)g of AgBr

For the second equation
x g of NaBr produces y g of AgBr

Notice that g (NaBr + CaBr2) = 0.9157-x +x = 0.9157g (just as the question says)

Also notice that the total g (AgBr) from both equations = 1.693-y+y g = 1.693g.

The key is to solve for x and y (2 equations, 2 unknowns).
Here's how you set it up:

From equation 1:
(0.9157-x)g CaBr2 * 1mole CaBr2/200g CaBr2 * 2mole AgBr/1mole CaBr2 * 188g AgBr/ 1mole AgBr = (1.693-y) g AgBr (1)

From equation 2:
x g (NaBr) * 1mole (NaBr)/103g (NaBr) * 1 mole (AgBr)/ 1mole (NaBr)
* 188g (AgBr)/I mole (AgBr) = y g AgBr (2)

If you break down the 2 equations, you get:
376x - 200y = 5.7 (3)
1.82x = y (4)

if you solve for x and y, you get
x = 0.475g
y = 0.865g


you can confirm this result by plugging in x and y into (1) and (2)

Since we know that x = g NaBr
Therefore g NaBr = 0.475g

% NaBr in the mixture
= 100 * 0.475/0.9157
= 51.87%

Hope this helps

solasoy

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