# Chemistry Question

Discussion in 'DAT Discussions' started by Ferdowsi, May 27, 2008.

1. ### Ferdowsi

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Hey all,

How many grams of chlorine gas are needed to produce 3 L of HCL gas at a pressure of 2 atm and a temparture of 19 C?

here is the equ: H2 + Cl2---> 2HCl

3. ### userah

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hmm okay let's see here. Based on all the information they give you, first calculate the number of moles you have of HCl using PV = nRT

2(3) = n(.08)(273 + 19)
6 = (approx) 24n
n = 6/24 or 1/4 or .25

then .25 mol HCL * (2mol Cl/ 2mol HCl) * (35gCl/1mol Cl)

giving you about 8.75grams of Cl

I think that's what it is...

4. ### doc toothache

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Convert 3L of HCl to stp, divide by 22.4 to gel moles of HCl, then divide by 2 to get number of moles of Cl2, then multiply by eq. wgt chloride to obtain grams.

5. ### shankin

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M.W.= [mass (x)*R*T]/[P*V] --> modified ideal gas equation

35(apx)= x*24(apx)/6

solve for x

mass= 8.75 (like userah) said

6. ### Zerconia2921 Bring your A-game!

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PV=nRT

n=PV/RT
n = 2atm(3L)/0.0821)*(273+19)
n=6/24
n=0.25moles of HCL
0.25molesHCL*1moleCL2/2molesHCl *70gCl2/1moleCl2 = 8.75grams of CL2

7. ### AFF2009

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I did it exactly like u did it and i still got 8.98.. could u double check. thanks

8. ### Zerconia2921 Bring your A-game!

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I miss multiplied R*T (.0821*292)=24.0
n=6/24.0
n=1/4
n=0.25moles HCl and the rest is just stoci giving you 8.75grams

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10. ### Ferdowsi

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y using STP for this problem? as far as the questions says, there is not standard involved? so y using STP.. I don't get it..

11. ### shankin

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292*.0821 is aprox 24 not aprox 240

you mis multiplied again

the answer is still 8.75 (aprox)

12. ### Zerconia2921 Bring your A-game!

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haha damn i got it right now my bad

13. ### Zerconia2921 Bring your A-game!

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make sure your units are correct and using the correct R value.

14. ### shankin

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"How many grams of chlorine gas are needed to produce 3 L of HCL gas at a pressure of 2 atm and a temparture of 19 C?"

Given:

here is the equ: H2 + Cl2---> 2HCl

since you're looking for gas you need to use a gas equation or gas law

since your given PV and T it's pretty easy to assume that you use the ideal gas equation PV=nRT

since your looking for mass in grams and not moles you have to adjust

moles= mass (grams)/MW

so solving for n you get the following

n= PV/RT moles

moles (in this case n) = Mass over MW

thus

Mass/Mw= PV/RT moles

you know the MW of Cl is approx 35
you know P is 2 atm
you know V is 3 L
you know r is .0821 a constant
you know T is 19 C or 19+273= 292K

so you have
mass/35= (2*3)/(292*.0821)

thus mass= 2*3*35/(292*.0821)

so mass= approx 8.759

if you use the real mass of Cl (35.4 ish) you get 8.86 ish

15. ### doc toothache

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In order to calculate the number of moles you have it needs to be compared with the volume at stp, or if it makes you feel better you can work backwards and calculate how much one mole (22.4 L)of gas will occupy at 2 atm and 19 C.

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