How many grams of HCl must be used to produce 10.0L of chlorine gas at STP?

Reaction: 2HCl (g)--->H2 (g)+Cl2 (g)

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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How many grams of HCL are required to produce 224 liters of CL2at stp

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