Determine the volume of hydrogen collected at stp and satp, and the molar volume at stp and satp

The other given data are: 
Room temperature (°C) 22.0; pressure (kPa) 100.5; Water vapor pressure at 22 °C (2.60 kPa); Mass of Mg ribbon (0.05g);  Volume of hydrogen gas (mL) 48.3

Vstp = 22.4 L/mol, and Vsatp = 24.8 L/mol Vstp:0 degrees Celsius and 101.325 kPa while Vsatp: 25 degrees Celsius and 100 kPa. 100.5 kPa x 760 mm Hg/101.325 kP = 753.8119911 mmHg
 
2HCl(g) + Mg(s) -> H2(g) + MgCl2(aq) nMg = 0.0020571899 mol nH2 = 0.0020571899 mol Wet H2 pressure = 100.5 kPa Dry H2 pressure = 97.9 kPa

So to determine the volume of hydrogen gas collected at STP: V2 = (753.8119911 mmHg)(48.3 mL)(273 K) / (295 K)(760 mmHg) = 44.33403003 mL.

Molar volume at STP: mol/L = 44.33403003 mL / 0.0020571899 mol
= 21550.77176 mL/mol
= 21.55077176 L/mol
To determine volume of hydrogen gas collected at SATP:
V2 = (100.5 kPa)(48.3 mL)(298 K) / (295K)(100kPa) = 49.03514237 mL

Molar volume at SATP: mol/L = 49.03514237 mL / 0.0020571899 mol
 = 23835.98246 mL/mol
= 23.8359824 L/mol

Answer; carbon monoxide (co);