Asked by Cliff

How many grams of HCl must be used to produce 10.0L of chlorine gas at STP?
Reaction: 2HCl (g)--->H2 (g)+Cl2 (g)
Answered by bobpursley
If you had 22.4 liters, that would be a mole of CL2, which required 2 moles of HCl

answer: gramsHCL= molesHCL*36g/mole
= 2molesHCL/moleCl2*10/22.4*molesCl2*36gramsHCL/moleHCL
=2*10/22.4 * 36 gHCl
Answered by Jorge
How many grams of HCL are required to produce 224 liters of CL2at stp
Answered by Petr Cech
Prolly take something and the other thing and ask it why it is. After that say thanks.
Answered by Manuel Nuer
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Answered by Gianluigi Buffon
Manuel I'm better than you at goalkeeper and Chemistry. You're not the best goalkeeper, I'm Italy's finest.
Answered by Iker Casillas
You are all forgetting one of the world's all time best Goalkeeper. It's me, Iker.
Answered by Brianna
You would use the ideal gas law: PV=nRT. V=10.0L T=273°K P=1.00atm R=.0821. You can rearrange the equation so it’s PV/RT=n. Which is (1.00x10.0)/(0821x273)=.446molCl2 you now need to convert .446molCl2 to grams of HCL. You use stoichiometry to do this. So it would be (.446molCl2x2molHCLx36.46gHCL)/(1.00molCl2x1molHCL) this will equal 32.6gHCL which is your final answer
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