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Topic: Volume of Gaseous Products  (Read 7398 times)

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Offline oceanmd

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Volume of Gaseous Products
« on: November 08, 2010, 10:49:02 PM »
Determine the volume of gaseous products (at STP) produced by the detonation of 2 (English) tons of ammonium nitrate (less than half of which is thought to have actually detonated.)

What are the steps to solve it?
Gaseous products are N2 and CO2. In the balanced equation it's 52 mol of N2 and 17 mol of CO2
Should I use the Ideal Gas Law? Then V=nRT/P, V= (52+17) x 8.314 x 273/101.3=1.5 x 10^3

This answer is wrong. Please tell me what I am doing wrong.

Thank you very much

Offline oceanmd

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Re: Volume of Gaseous Products
« Reply #1 on: November 09, 2010, 01:40:31 AM »
I got it. I need to find how many moles there are in 1 ton of ammonium nitrate first, then the rest is cleart.

Thank you   

Offline Borek

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Re: Volume of Gaseous Products
« Reply #2 on: November 09, 2010, 03:02:42 AM »
Gaseous products are N2 and CO2

I can't decide whether to cry or to laugh.
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Offline oceanmd

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Re: Volume of Gaseous Products
« Reply #3 on: November 09, 2010, 02:20:09 PM »
Why? They are both gases.

Offline Borek

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Re: Volume of Gaseous Products
« Reply #4 on: November 09, 2010, 04:16:24 PM »
SO2 is a gas too.
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Offline sjb

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Re: Volume of Gaseous Products
« Reply #5 on: November 09, 2010, 05:20:27 PM »
What's the balanced equation for the decomposition? 52/17 is a very odd ratio of products and ammonium nitrate isn't that complex.

Offline Borek

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Re: Volume of Gaseous Products
« Reply #6 on: November 09, 2010, 06:20:12 PM »
52/17 is a very odd ratio of products

Odd ratio of odd products.
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Offline oceanmd

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Re: Volume of Gaseous Products
« Reply #7 on: November 09, 2010, 07:11:39 PM »
Possible expression for the chemical reaction that destroyed Murrah Building in Oklahoma City in 1955
52NH4NO3 + C17H36 = 52N2 +17CO2 + 122H2O

I thought that the problem about the detonation of 2 English tons of ammonium nitrate is based on the above reaction.

Offline oceanmd

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Re: Volume of Gaseous Products
« Reply #8 on: November 09, 2010, 09:02:39 PM »
OK, now more on this problem.
1. we will assume that 1 ton reacted. 1 ton = 2240 lb x 454g =1016960g
2. 1 mol of NH4NO3 is 80 g, so 1016960 g is 12712 mol
3. According to this equation 52 NH4NO3 + C17H36 = 52 N2 + 17 CO2 + 122 H2O
   52 mol of NH4NO3 produces 52+17+122=191 moles, then 12712 mol produces 46692 mol
4. 1 mol = 22.4L, so 46692 mol = 1045900 L = 1.04 x 10^6

Please advise if this makes sense.

Thank you

Offline Borek

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Re: Volume of Gaseous Products
« Reply #9 on: November 10, 2010, 04:22:03 AM »
Possible expression for the chemical reaction that destroyed Murrah Building in Oklahoma City in 1955
52NH4NO3 + C17H36 = 52N2 +17CO2 + 122H2O

I thought that the problem about the detonation of 2 English tons of ammonium nitrate is based on the above reaction.

This is not explosion of ammonium nitrate, this is explosion of ammonium nitrate mixed with heptadecane.

So, do you have 2 tons of the mix, or two tons of ammonium nitrate in the mix?
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Offline oceanmd

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Re: Volume of Gaseous Products
« Reply #10 on: November 10, 2010, 12:51:39 PM »
his is the original problem
Determine the volume of gaseous products (at STP) produced by the detonation of 2 (English) tons of ammonium nitrate (less than half of which is thought to have actually detonated.) And this is the given equation 52NH4NO3 + C17H36 = 52N2 +17CO2 + 122H2O

Thank you

Offline Borek

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Re: Volume of Gaseous Products
« Reply #11 on: November 10, 2010, 12:55:12 PM »
Question is ambiguous. Not your fault.
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