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Chapter 1: Stoichiometry

Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Fri Jul 01, 2016 2:57 pm

Bellesp wrote: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?


Sorry about that!

For reasons that I can't explain, the superscript + is not showing between "H" and the period:

H .

should be

H^+.

It is in the Admin but not on the live webpage for reasons that I can't explain. I'll have a programmer look into it.

Side note: "H" all by itself cannot exist in a normal chemical reaction because it is way too unstable. You'll almost always see H (the hydrogen atom) as part of a molecule (i.e. H2, H2O, H2SO4) or as the hydrogen cation (H^+).
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Re: Chapter 1: Stoichiometry GAMSAT level Q4

Postby natasha.mi3015 » Mon Aug 12, 2019 11:17 am

Hi, I am from a non-science background and at the beginning of my studies so I am probably a bit slow picking up on "in between" steps in the explanation to a solution; I apologise in advance but I am finding it difficult to follow the explanation of the solution.

"4) How much methane can be generated if the lab processes 12 kg H2 with 79.8 kg Fe2O3 and 132 kg CO2?

Explanation: Converting weights to numbers of moles per compound, we have; H2 = 6 moles; 12/(1 x 2) = 6. CO2 = 3 moles; 132/44 = 3. Fe2O3 = 1/2 mole, 79.8/159.6 = 0.5. Hydrogen and Carbon exist in equal proportion. 3 moles C + 6 moles H2 (3 moles 2 H2) react to form 3 moles CH4. 3 moles CH4 weigh 3 x (12 + 4) = 48 kg. "

- How do you find that 3 moles of C + 6 moles of H2 produce 3 moles of CH4?
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Mon Aug 12, 2019 4:27 pm

natasha.mi3015 wrote:Hi, I am from a non-science background and at the beginning of my studies so I am probably a bit slow picking up on "in between" steps in the explanation to a solution; I apologise in advance but I am finding it difficult to follow the explanation of the solution.


No apologies necessary, you should be congratulated for your efforts!

natasha.mi3015 wrote:- How do you find that 3 moles of C + 6 moles of H2 produce 3 moles of CH4?


The passage provides the balanced reaction:

(iii) Creation of Methane:

2H2 + C → CH4

If we can figure out the number of moles of H2 and C, then stoichiometry (the balanced reaction) allows us to conclude the number of moles methane CH4. Notice that the number of moles in the reaction (iii) above can be represented as: 2:1:1 (meaning H2:C:CH4).

The Explanation shows how weight is converted to the number of moles. Some students like to memorize that the number of moles n = (weight)/(molecular weight), this can also be represented (dimensional analysis) as: moles = (grams)/(grams/mole) which shows that 'grams' cancel leaving the answer in moles.

From the Explanation, we know that there are 6 moles of H2 and 3 moles of CO2. But CO2 is not in equation (iii) so we need to find a relationship between CO2 and C. The preceding is found in equation (ii) where the relationship between the 2 molecules can be represented as 1:1, so in this case, 3:3. Now we know that 6 moles of H2 reacts with 3 moles of C. Since the ratio of (iii) is represented as 2:1:1 (meaning H2:C:CH4), multiplying through be 3 we get: 6:3:3. So 6 H2 react with 3 C to make 3 CH4.

Since: moles = (grams)/(grams/mole)

Then: moles(grams/mole) = (grams) (notice how 'moles' cancels)

So for methane: 3 (16) = 48


You may want to watch (or re-watch!) the video Stoichiometry for Chapter 1 (also found in the Videos section of gamsat-prep.com). Yes, it does get easier as you complete more practice questions!

Please feel open to asking further questions to clarify what you encounter.
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Re: Chapter 1: Stoichiometry

Postby zk_clo » Mon Feb 17, 2020 10:18 am

Hi there,
Chap 1 GAMSAT level MCQs, q5. I don't see how "9 kg H2O is equivalent for our discussion to 1/2 mole H2O" (It says I got it right, but that was luck!)
Thanks!

5) Given only 9 kg of H2O, what is the maximum amount of methane that the lab can produce from the rover's waste products?
A) 1 kg
B) 2 kg
C) 4 kg
D) 8 kg

Result: Your answer is correct.
Your choice: C Correct choice: C

Explanation: 9 kg H2O is equivalent for our discussion to 1/2 mole H2O, which yields 1/2 mole H2 by equation (iv), and so can produce 1/4 mole CH4 by equation (iii). 1/4 mole CH4 weighs 1/4 (16 kg) = 4 kg.

Please note: since all the masses are in kilograms (including the answer), there is no need to convert to grams. Of course, you can choose to do so but it would take more time.
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Tue Feb 18, 2020 5:53 am

One mole of H2O is 18 g. 18 kg (10^3 grams) would be 1000 moles and so 9 kg is 500 moles. It's perfectly fine to work with those numbers, but if there are no conversions needed to other SI units and if we are only dealing with ratios, we can just 'pretend' that we are working with grams and then remember at the end that we were really working with kilograms.
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Re: Chapter 1: Stoichiometry

Postby HunterEd » Wed Apr 22, 2020 1:12 am

Hello,

Need help please!

First attempt at GenChem so forgive me for asking a question that has already been answered. I've read over the explanation a million times yet I still can't seem to work it out :(

5) Given only 9 kg of H2O, what is the maximum amount of methane that the lab can produce from the rover's waste products?
A) 1 kg
B) 2 kg
C) 4 kg
D) 8 kg

Explanation: 9 kg H2O is equivalent for our discussion to 1/2 mole H2O, which yields 1/2 mole H2 by equation (iv), and so can produce 1/4 mole CH4 by equation (iii). 1/4 mole CH4 weighs 1/4 (16 kg) = 4 kg.

9kg of H2O is equal to 1/2 mole H20 (9kgs/18 g/mol), this part I get...however can't seem to workout the below.

"which yields 1/2 mole H2 by equation (iv), and so can produce 1/4 mole CH4 by equation (iii). 1/4 mole CH4 weighs 1/4 (16 kg) = 4 kg"

Thank you,
Hunter.
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Wed Apr 22, 2020 3:05 pm

I'm not exactly sure what part of the solution is at issue, please have a look at the following and then let me know if something is still missing:

viewtopic.php?f=6&t=3356

(never worry about asking questions!)
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Re: Chapter 1: Stoichiometry Questions

Postby SeoM » Sun May 03, 2020 2:17 am

Q2. In this question, it does not give you the atomic mass of Mn / CL2. Would you be expected to know these in the exam??
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Sun May 03, 2020 1:46 pm

SeoM wrote:Q2. In this question, it does not give you the atomic mass of Mn / CL2. Would you be expected to know these in the exam??


ACER expects that you know the trends in the periodic table, very rarely they would have a question where you would have been expected to know the amu of a very common atom (i.e. C, H, O or N) but not Cl or Mn. We have now added the amu's to the end of the passage.
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Chapter 1: Stoichiometry - Question 16

Postby dosani.rk7379 » Mon Feb 01, 2021 9:11 pm

Hi,

I am a bit confused about question 16 of Chapter 1: Stoichiometry from the new Masters Series General Chemistry textbook.

I understand how to calculate the moles per compound, however, I'm not sure how to use the calculated moles to determine the kg of methane generated. Which chemical equation do we follow?

Thanks!
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Re: Chapter 1: Stoichiometry

Postby goldstanda3269 » Tue Feb 02, 2021 5:32 pm

dosani.rk7379 wrote:Hi,

I am a bit confused about question 16 of Chapter 1: Stoichiometry from the new Masters Series General Chemistry textbook.

I understand how to calculate the moles per compound, however, I'm not sure how to use the calculated moles to determine the kg of methane generated. Which chemical equation do we follow?

Thanks!


We want methane as a product so only reaction (iii) permits that. Now we examine: What makes methane? We have hydrogen (and we know the quantity so that is helpful). And we have carbon but we do not have that quantity directly, but we need to know it.

OK, so how to we make carbon? Reaction (ii) can make carbon and fortunately we are given the quantity of the reactant CO2 which can lead us to the quantity of carbon C.

Now we know the amount of C and H so we can determine CH4 with reaction (iii).

And now, we can see that 79.8 kg of Fe2O3 was a distractor ("a red herring").
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