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Author Topic: Molar Volume and Stoicheometry  (Read 590 times)

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DessaM

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Molar Volume and Stoicheometry
« on: November 25, 2012, 09:34:38 PM »

I'm trying to figure out how to set up this equation. Each time I seem to be getting the wrong answer.

Zinc reacts with aqueous sulfuric acid to form H gas: Zn + H2SO4 -----> ZnSO4 + H2
What mass of Zinc is necessary to make 24.9L of hydrogen gas at 1.44atm and 508.15K ?

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Borek

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Re: Molar Volume and Stoicheometry
« Reply #1 on: November 25, 2012, 10:15:54 PM »

Show how you get the wrong answer, we will start from there.
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DessaM

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Re: Molar Volume and Stoicheometry
« Reply #2 on: November 25, 2012, 10:25:13 PM »

Show how you get the wrong answer, we will start from there.

This is how I had set it up the first time

22.4 L H2 * (1 mol H2/ 352.88 L H2) * (1 mol Zn/ 1 mol H2) * (65.39g Zn/ 1 mol Zn) = 4.614 g Zn

I know the (1 mol H2/ 352.88 L H2) is incorrect. That's where I'm having the most difficulty is how to set up equations that don't involve STP.
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sjb

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Re: Molar Volume and Stoicheometry
« Reply #3 on: November 25, 2012, 10:30:14 PM »

Show how you get the wrong answer, we will start from there.

This is how I had set it up the first time

22.4 L H2 * (1 mol H2/ 352.88 L H2) * (1 mol Zn/ 1 mol H2) * (65.39g Zn/ 1 mol Zn) = 4.614 g Zn

I know the (1 mol H2/ 352.88 L H2) is incorrect. That's where I'm having the most difficulty is how to set up equations that don't involve STP.

Where has the 352.88 figure come from?
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DessaM

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Re: Molar Volume and Stoicheometry
« Reply #4 on: November 25, 2012, 10:36:06 PM »

Show how you get the wrong answer, we will start from there.

This is how I had set it up the first time

22.4 L H2 * (1 mol H2/ 352.88 L H2) * (1 mol Zn/ 1 mol H2) * (65.39g Zn/ 1 mol Zn) = 4.614 g Zn

I know the (1 mol H2/ 352.88 L H2) is incorrect. That's where I'm having the most difficulty is how to set up equations that don't involve STP.

Where has the 352.88 figure come from?

I got that from v=t/p (which I know is incorrect) I just did v=508.15 / 1.44atm = 352.88  My teacher just crossed it out on my worksheet and hasn't shown me the correct way yet..
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DrCMS

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Re: Molar Volume and Stoicheometry
« Reply #5 on: November 25, 2012, 11:09:43 PM »

The equation you need is the ideal gas law PV = nRT http://en.wikipedia.org/wiki/Ideal_gas_law

You have tried to use a bit of the equation (v=t/p) but it only works if you use the full equation.

You need to pick the right value of R (http://en.wikipedia.org/wiki/Ideal_gas_constant) for the units you have been given for P, V and T and solve for n. 
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DessaM

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Re: Molar Volume and Stoicheometry
« Reply #6 on: November 26, 2012, 12:33:45 AM »

The equation you need is the ideal gas law PV = nRT http://en.wikipedia.org/wiki/Ideal_gas_law

You have tried to use a bit of the equation (v=t/p) but it only works if you use the full equation.

You need to pick the right value of R (http://en.wikipedia.org/wiki/Ideal_gas_constant) for the units you have been given for P, V and T and solve for n.

Okay so I applied the idea gas law using   n=PV/RT so I got .859 mol H2

Then I did this:

 .859 mol H2 * (1 mol Zn / 1 mol H2) * (65.39 g Zn / 1 mol Zn) = 56.17g Zn

Is that the correct way to set up the equation? And am I using the mol ratios correctly?
Thank you for your *delete me*
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curiouscat

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Re: Molar Volume and Stoicheometry
« Reply #7 on: November 26, 2012, 01:21:52 AM »

The equation you need is the ideal gas law PV = nRT http://en.wikipedia.org/wiki/Ideal_gas_law

You have tried to use a bit of the equation (v=t/p) but it only works if you use the full equation.

You need to pick the right value of R (http://en.wikipedia.org/wiki/Ideal_gas_constant) for the units you have been given for P, V and T and solve for n.

Okay so I applied the idea gas law using   n=PV/RT so I got .859 mol H2

Then I did this:

 .859 mol H2 * (1 mol Zn / 1 mol H2) * (65.39 g Zn / 1 mol Zn) = 56.17g Zn

Is that the correct way to set up the equation? And am I using the mol ratios correctly?
Thank you for your *delete me*
I'm getting the same answer. So either we are both right or both wrong. :)
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DrCMS

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Re: Molar Volume and Stoicheometry
« Reply #8 on: November 26, 2012, 02:34:08 AM »

Watch out for rounding up part way through I get 0.8599... moles of hydrogen coming from 56.23g of Zinc. 
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