3.5 cm of Mg ribbon(0.009729 g/cm0 reacts with HCL to produce 32.6 ml. of Hydrogengas at 750 mm Hg and 20 degree C. the vapor pressure of water at 20 degree C is 17.5 mm Hg. Calculate the molar volume of hydrogen (L/mole) and the percent error in your determination

Answer 1

I suppose you mean the mass of Mg = 0.009729 g/cm.
Mass Mg = 0.009729g/cm x 3.5 cm = approximately 0.034g (but you can get a more accurate answer).
mols Mg = grams/molar mass = 0.034/24.3 = 0.0014 = mols H2 (from the equaton below.)

Mg + 2HCl ==> H2 + MgCl2

32.6 mL wet H2 gas at the conditions listed. Convert this to STP for DRY H2 gas. Use (P1V1/T1) = (P2V2/T2)
P1 = 750-17.5 = ?mm
V1 = 32.6 mL
T1 = 273+20 = ?
P2 = 760 mm
V2 = ?
T2 = 273
I get V2 = approximately 30 mL and that is for 0.0014 mol.
For 1 mol that will be 30/0.0014 = about 20,000 mL/mol. The recognized molar volume is 22,400 mL/mol. Your experimental error is
[(20,000 - 22,400)/(22,400)]*100 = about 6%