Chemistry Question

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Ferdowsi

Full Member
10+ Year Member
15+ Year Member
Joined
May 25, 2008
Messages
84
Reaction score
0
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

your feedback will be appreciated:hardy:

Members don't see this ad.
 
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...
 
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

your feedback will be appreciated:hardy:

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.
 
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
 
Members don't see this ad :)
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

your feedback will be appreciated:hardy:

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
 
PV=nRT

n=PV/RT
n = 2atm(3L)/0.0821)*(273+19)
n=6/23
n=0.3moles of HCL
0.3molesHCL*1moleCL2/2molesHCl *70gCl2/1moleCl2 = 10.5 grams of CL2


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

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
 
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.


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..:confused:
 
292*.0821 is aprox 24 not aprox 240

you mis multiplied again

the answer is still 8.75 (aprox)
 
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..:confused:

your question asked


"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 :thumbdown: 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
 
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..:confused:

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.
 
your question asked


"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 :thumbdown: 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

This all makes perfect sense, but 1 question. For me, the math is the most difficult part so, don't laugh.....but..........how do you compute the last formula (292 x 0.0821) without a calculator!!!! Ahh if it wasn't for math.
 
Top