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?

Answer 1

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 )