What volume of HCl gas is required to react with excess magnesium metal to produce 6.82 L of hydrogen gas at 2.19 ATM and 35 degrees C?

Balanced equation first Mg + 2HCl >> MgCl2 + H2 ( use PV = nRT to find moles H2 ) ( C to K ) (1atm)(6.82L) =n(0.08206Latm/molK)(308.15K) 0.2697 moles H2 ( drive back against HCL to find those moles ) 0.2697 moles H2 (2moles HCl/1mole H2) 0.5394 moles HCl ( conditions remain the same, so use PV = nRT to find volume HCl ) (1atm)(Vol) = (0.5394 moles HCl)(0.08206Latm/molK)(308.15K) 13.6 Liters HCl gas required ( may be faster way, but it is many years since I have done any chemistry ).

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