# chemistry

How many grams of HCl must be used to produce 10.0L of chlorine gas at STP?
Reaction: 2HCl (g)--->H2 (g)+Cl2 (g)

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1. If you had 22.4 liters, that would be a mole of CL2, which required 2 moles of HCl

= 2molesHCL/moleCl2*10/22.4*molesCl2*36gramsHCL/moleHCL
=2*10/22.4 * 36 gHCl

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

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3. Prolly take something and the other thing and ask it why it is. After that say thanks.

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4. Petr I knew I'd find you here. I'm the worlds best goalkeeper.

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5. Manuel I'm better than you at goalkeeper and Chemistry. You're not the best goalkeeper, I'm Italy's finest.

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6. You are all forgetting one of the world's all time best Goalkeeper. It's me, Iker.

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7. 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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