Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: DessaM on November 26, 2012, 03:34:38 AM

Title: Molar Volume and Stoicheometry
Post by: DessaM on November 26, 2012, 03:34:38 AM
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 ?

Title: Re: Molar Volume and Stoicheometry
Post by: Borek on November 26, 2012, 04:15:54 AM
Show how you get the wrong answer, we will start from there.
Title: Re: Molar Volume and Stoicheometry
Post by: DessaM on November 26, 2012, 04:25:13 AM
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.
Title: Re: Molar Volume and Stoicheometry
Post by: sjb on November 26, 2012, 04:30:14 AM
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?
Title: Re: Molar Volume and Stoicheometry
Post by: DessaM on November 26, 2012, 04:36:06 AM
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..
Title: Re: Molar Volume and Stoicheometry
Post by: DrCMS on November 26, 2012, 05:09:43 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. 
Title: Re: Molar Volume and Stoicheometry
Post by: DessaM on November 26, 2012, 06:33:45 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*
Title: Re: Molar Volume and Stoicheometry
Post by: curiouscat on November 26, 2012, 07:21:52 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. :)
Title: Re: Molar Volume and Stoicheometry
Post by: DrCMS on November 26, 2012, 08:34:08 AM
Watch out for rounding up part way through I get 0.8599... moles of hydrogen coming from 56.23g of Zinc.