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### Topic: Homework problems that completely stump me  (Read 15747 times)

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#### britbro19

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##### Homework problems that completely stump me
« on: September 20, 2008, 05:42:15 PM »
I need some help with 3 homework questions. We're really behind due to missing so much school because of Hurrican Gustav and Hurricane Ike; therefore, I'm having a hard time catching up. Any help would be greatly appreciated.

1. Calculate the percent yield of an analysis if 1.9365g of AgCl (MM 143.32g/mol) was precipitated in a gravimetric experiment if 1.2000g of a soluble sample that contained 40.25% Cl- was used as the starting material.

--Ok, I know for gravimetric analysis you need to know the starting material and ending material which would be 1.2g of sample and 1.9365 g AgCl product. After this step, I'm lost.

2.A mixture containing only Al2O3 (MM 101.96 g/mol) and Fe2O3 (MM 159.69 g/mol) weighs 2.0190g. When heated under a stream of H2, Al2O3 is unchanged, but Fe2O3 decomposes to Fe(s) and water vapor. If the residue weighs 1.7740g, what is the %Fe2O3 in the original sample?

Fe2O3--->2Fe(s) + 3H20(g)

Hint: Think about what's left in the residue and where the change in mass comes from!

--This one has me 100% lost. I don't even know where to start.

3. When a 100.0mL portion of a solution containing 0.5000g of AgNO3 (MM 169.87 g/mol) is mixed with 100.0 mL of a solution containing 0.3000g of K2CrO4 (MM 194.19 g/mol), a bright red precipitate of Ag2CrO4 (MM 331.73 g/mol) forms. Calculate the molar concentration of the excess reactant that remains in solution.

--I have no idea what to do if there's 2 starting materials.

Thanks for all of your time

#### Borek

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##### Re: Homework problems that completely stump me
« Reply #1 on: September 20, 2008, 06:09:33 PM »
1. How many moles of Cl- in the starting sample? How much AgCl should be precipitated with that amount of chlorides?

2. Hint you were given is a very good one. You started with Fe2O3/Al2O3 mixture. What is left in the mixture after the reaction?

3. This is a limiting reagent question. Start with the reaction equation.
http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents
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#### britbro19

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##### Re: Homework problems that completely stump me
« Reply #2 on: September 20, 2008, 06:24:00 PM »
Ok, for #1 I know there are 0.013512 mol Cl-...I'm still lost after that.

I'm still lost for the rest, I know I'm really slow in chemistry, sorry.

#### Borek

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##### Re: Homework problems that completely stump me
« Reply #3 on: September 20, 2008, 06:57:02 PM »
Ok, for #1 I know there are 0.013512 mol Cl-

Write balanced reaction equation.

Do you know how to read these equations? Check here:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

Now, assuming you started with 0.013512 moles of Cl-, how many moles of AgCl should precipitate? What mass should it have?

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#### britbro19

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##### Re: Homework problems that completely stump me
« Reply #4 on: September 20, 2008, 07:04:04 PM »
OK:

Ag     +     Cl     --->   AgCl
.4025 g       1.9365g

Is this right? I convert the percent they gave me to grams? Or do I put them both in moles?

Ag    +      Cl     ----->    AgCl
.013512mol        .013512mol

Or is this right?

Then what do I do?

#### Borek

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##### Re: Homework problems that completely stump me
« Reply #5 on: September 20, 2008, 07:10:52 PM »
Ag    +      Cl     ----->    AgCl
.013512mol        .013512mol

This is how the reaction proceeds - you should end with 0.013512 mole of AgCl. Now, calculate its mass - that'll be theoretical yield. You know real yield (that was given) - use both numbers to calculate percent yield.
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#### britbro19

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##### Re: Homework problems that completely stump me
« Reply #6 on: September 20, 2008, 11:22:47 PM »
Wait...if I change 0.013512 moles of AgCl to grams of AgCl that gives me 1.9365 g of AgCl, which is the same number the problem gives...so that'd be 100% yield.

I think I'm misunderstanding what you said. Can you explain it in a different way. What two numbers am I using for percent yield?  Do I even need to use the 40.25% Cl- in the problem?

#### Borek

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##### Re: Homework problems that completely stump me
« Reply #7 on: September 21, 2008, 04:37:33 AM »
Pretty close, but on the closer scrutiny numbers you list are a little bit strange.

Where did you get 0.013512 from?

What two numbers am I using for percent yield?

percent yield = practical yield/theoretical yield * 100%

Quote
Do I even need to use the 40.25% Cl- in the problem?

Yes, to calculate theoretical yield.
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#### britbro19

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##### Re: Homework problems that completely stump me
« Reply #8 on: September 21, 2008, 01:59:01 PM »

1.9365 g AgCl  x 1 mol of AgCl  x  1 mol Cl
143.321 g AgCl    1mol AgCl   = 0.013512 mol Cl

What's the formula I need to use to find the percent yield? And where does that 40.25% come into play?

I did figure out how to do #3 though, YAY!

#### britbro19

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##### Re: Homework problems that completely stump me
« Reply #9 on: September 21, 2008, 02:04:31 PM »
I'm sorry, I meant what's the formula to find theoretical yield?

#### Borek

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##### Re: Homework problems that completely stump me
« Reply #10 on: September 21, 2008, 02:08:10 PM »
40.25% is for calculation of the MASS of chlorides.

Then you are trying to convert this MASS of chlorides to MOLES of chlorides. OK. But you don't use molar mass of AgCl for that.

Theoretical yield is the amount you get from stoichiometric calculations. You know mass/moles of chlorides, calculate how much AgCl should be produced if chlorides react completely. That will be your theoretical yield.
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#### britbro19

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##### Re: Homework problems that completely stump me
« Reply #11 on: September 21, 2008, 02:18:24 PM »
How do I use 40.25% to find the mass of chlorides?

.4025 mol Cl x 35.452 g Cl
1 mol Cl         = 14.27 g Cl

Is this correct?

#### Borek

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##### Re: Homework problems that completely stump me
« Reply #12 on: September 21, 2008, 02:31:01 PM »
You are told that 1.2 g of the sample contains 40.25% percent of chlorides. What does it mean?
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#### britbro19

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##### Re: Homework problems that completely stump me
« Reply #13 on: September 21, 2008, 02:43:53 PM »
I have no clue. I'm so confused

.4025 g Cl
1.2 g sample X 100 = 33.54 % Cl in sample?

#### Borek

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##### Re: Homework problems that completely stump me
« Reply #14 on: September 21, 2008, 02:54:57 PM »
Stop guessing blindly.

What is percentage definition?
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