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Chapter 1: Stoichiometry

Chapter 1: Stoichiometry

Postby admin » Fri Aug 27, 2010 12:40 am

Be the first to discuss this chapter!
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Percentage mass

Postby Emmalondon88 » Tue Oct 11, 2011 8:10 pm

Hi, I am having some trouble understanding one of the problems in section 1.4 Percentage Mass. In the second problem we are given the weight of vitamin C as 4.00mg which yields 6.00mg CO2 and 1.632mg H2O. I'm not sure if I'm just missing something, but how is 4mg yielding 7.632mg? Thanks for your help.
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Postby goldstanda3269 » Thu Oct 13, 2011 11:18 pm

Good question.

You are really asking: what happened to the conservation of mass? Shouldn't the total mass in the reactants equal the total mass in the products?

The answer is, unless it is a nuclear reaction (Physics Chapter 12), YES !!!

So what happened?

A fireman would say: no oxygen, no fire. The problem says that the Vit C was burning which suggests unlimited oxygen. That is exactly the missing mass (Combustion is discussed on Organic Chem chapter 3).
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Re: Chapter 1: Stoichiometry

Postby Tmeredith7702 » Sat Jun 14, 2014 4:03 am

The lesson video for this chapter doesn't seem to be the right one (it talks about "Physic" and up and down arrows). Is that right or where is the correct video lesson please?
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Sun Jun 15, 2014 4:59 am

It is the correct video. In the first 3 minutes, Dr. F is answering a student's question about assessing the sign for gravity. You can ignore that and watch the video from the 3 minute mark.
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A question about (8) in Q & A

Postby li.zhang96890 » Mon Oct 05, 2015 3:55 pm

"8) Consider the following reaction:
FeCl2(aq) + H2S(g) --> FeS(s) + 2HCl(aq)

When sulfur is precipitated, what type of reaction has occurred?


A) Oxidation-reduction
B) Neutralization
C) Double replacement
D) Displacement

Result: Your answer is incorrect.
Your choice: D Correct choice: C"

I want to know if the displacement reaction is also called single- replacement reaction, in which one element is substituted for another element in a compound, generating a new element and a new compound as products.
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Tue Oct 06, 2015 8:08 am

The reaction shown in the question is a double replacement.

You did indeed describe a single replacement.

https://en.wikipedia.org/wiki/Single_di ... t_reaction
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Re: Chapter 1: Stoichiometry

Postby laura.cp897117 » Thu Oct 15, 2015 11:50 am

I'm having trouble with 1.4 Composition of a Compound by % Mass as well...

6mg yielded of CO2 * (12mg C / 44mg CO2) = 1.636mg of C in 6mg CO2 or 4mg Vit C
1.632mg yielded of H2O * (2.02mg H / 18mg H2O) = 0.183mg of H in 1.632mg H2O or 4mg Vit C
Therefore, mass of O in 4mg of Vit C = 4.00mg - 1.636 - 0.183 = 2.181mg of O in 4mg Vit C

C : H : O
1.636 : 0.183 : 2.181
divided by smallest (0.183)
8.9 : 1 : 11.9
9 : 1 : 12 = C9HO12 which is very different than C3H4O3!

Did I do something wrong here??
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Fri Oct 16, 2015 6:52 am

Your calculation was trying to go too far. We were only trying to get students to do very basic calculations to ensure that there is a fundamental understanding of the process. What you were trying to do is more advanced.

For example, let's first try to write what is happening (unbalanced combustion reaction):

Vit. C + O2 ----> CO2 + H2O

Notice that in CHM 1.4, we only discussed how you can calculate the percentage of carbon and hydrogen. We did not discuss oxygen because that would distract from the purpose of this section and include the complexity of the additional reactant. Now from the reaction above, you can see that because there are oxygen atoms in both reactants, the calculation is more complex (notice that your calculation had relatively too much oxygen because it considered Vit C as the only reactant: C9HO12).

Also, there is an issue with the way you calculated the amount of oxygen. Conservation of mass suggests that the total mass on the left equals the total on the right: 4 mg + X = 6 mg + 1.632 mg.
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Re: Chapter 1: Stoichiometry

Postby laura.cp897117 » Fri Oct 16, 2015 9:48 pm

That makes a lot more sense! Thank you!!!
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Re: Chapter 1: Stoichiometry

Postby tezzidy166196 » Thu Nov 05, 2015 2:03 pm

Hi Guys,

In the stoichiometry review questions, could you please explain how the oxidation number for Re was worked out to be 7 ?? i did work out the Re to be the one oxidised and NO3- to be reduced but the oxidation numbers given confuse me? Please help Thank-You :)


Here is the question and the explanation below ??

4) ReO4- + NO + H2O --> Re + NO3- + H . What are the oxidizing and reducing agents in the preceding chemical reaction, respectively?



A. ReO4- , NO
B. ReO4- , H2O
C. NO , ReO4
D. NO , H2O


Result: Your answer is incorrect.
Your choice: N/A Correct choice: A

Explanation: On the left of the equation, the oxidation numbers of Re and N are: Re (4 x 2) = -1 thus Re = 7

N (-2) = 0 thus N = 2

On the right of the equation: Re = 0

N (3 x 2) = -1 thus N = 5

As a result Re gained electrons as it went from 7 to 0 meaning Re is reduced = oxidizing agent (i.e. by being reduced it oxidizes the other compound). Conversely, N lost electrons (from 2 to 5) which is oxidation = reducing agent.
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Fri Nov 06, 2015 12:28 am

tezzidy166196 wrote:Hi Guys,

In the stoichiometry review questions, could you please explain how the oxidation number for Re was worked out to be 7 ??


ReO4-

Oxygen is -2 and there are 4 oxygens in ReO4- for a total of -8.

The total charge of ReO4- is -1.

Let the charge of Re = X.

X + (-8) = -1

X = 7
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Re: Chapter 1: Stoichiometry

Postby ElleRT » Mon Jun 06, 2016 5:00 pm

Hi,

Under 1.5 Calculating theoretical yield, why isn't the value of HCL at the end doubled? As at the end of the day we get 6HCL and not 3HCL. Thanks.
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Sat Jun 11, 2016 5:56 am

ElleRT wrote:Hi,

Under 1.5 Calculating theoretical yield, why isn't the value of HCL at the end doubled? As at the end of the day we get 6HCL and not 3HCL. Thanks.


Actually, we don't get 6 HCl. Why? Because we were given only 1 mole of BiCl3. If we had 2 moles of BiCl3 then we could make 6 moles of HCl according to the balanced reaction. But with only 1 mole of BiCl3, using a simple ratio that 2:6 is the same as 1:3, we produce only 3 moles of HCl.
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Q4 Oxidising agent

Postby Bellesp » Fri Jul 01, 2016 12:55 pm

In question 4, why is the hydrogen not also a reducing agent?

My understanding is that in the reactant H2O, H is 1+ and O is 2-
In the products, H is neutral so has an oxidation number of 0.

Doesn't this mean that H has reduced and is therefore an oxidising agent?

I'm not sure where I've gone wrong.

Thanks!
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