Calculate the volume of "dry" hydrogen that would be produced by one mole of magnesium at standard temperature and pressure.
Mg+2HCl=MgCl2+H2

The volume of H2 = 243mL = 0.243 L

Pressure of H2 gas collected = 745 mm

Pressure of H2 dry gas = P of H2 collected - vapor pressure of water = 745- 23.78 = 721.22 mm

= 721.22mm /760mm/atm

Temperature = 25C = 25+273= 298 K

Using the ideal gas equation

PV = nRT

or PV = (mass/molar mass) RT

(721.22/760)atm x 0.243L = (mass/ 2g/mol) x 0.0821L.atm/mol.K x 298K

Thus mass of H2 formed = 0.0188 g

Explanation:

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