Chemistry Question
Started by Ferdowsi
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...
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
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.
Advertisement - 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
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
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
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..😕
haha damn i got it right now my bad
make sure your units are correct and using the correct R value.
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 👎 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
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.
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.
