need help with #8, please show work. Thank you Watch a video of the procedure by clicking here. From the video you will need to record thefollowing data:Volume of H, gas collected in mL:42.95 mLHeight of the water column in cm:9.3 cmTemperature of water in °C:23,2°CPerform the following calculations:1. Using the balanced chemical equation below, determine the moles of hydrogen gasproduced when 0.0408 g Mg reacts.Mg + 2HC1 → MgCl, + H,2. Convert the height of the water column to the pressure of the water column in mm Hgusing the equation below(cm water)x10water13.63. Determine the pressure of the hydrogen gas if the air pressure was 751.5 mmHg.= PP H2– Pairwaer4. Convert the pressure of hydrogen gas from mmHg to atmospheres.mmHgatm =7605. Convert the temperature from Celceus to Kelvin.6. Determine the volume of hydrogen gas in L7. Use the ideal gas law, PV = nRT, to determine the value of the R.8. Based on a value of 0.0821 Latm/molK, determine the percent error for this experiment.experimental value of R-0.08210.0821x 100Percent Error =

1. Given that, Mass of Mg = 0.0408 g Molar mass of Mg = 24.305 g/mol moles of Mg = (0.0408g / 24.305g/mol) = 1.6786x10-3 mol According to the reaction stoichiometry, 1 mole of Mg produces 1 mol of H2 So, 1.6786x10-3 mol of Mg produces 1.6786x10-3 mol H2 moles of hydrogen produced = 1.6786x10-3 mol2. Given that, Height of the water column in cm = 9.3 cm Pwater = 9.3x1013.6 = 6.8382 mm Hg3. Pair = 751.5 mm Hg PH2 = Pair - Pwater PH2 = 751.5 - 6.8382 = 744.6618 mm Hg 4. atm = mm Hg760 = 744.6618760 = 0.9798 atm 5. Temperature = 23.2oC = 23.2 + 273.15 = 296.35 K 6. Volume of hydrogen gas collected in mL = 42.95 mL = (42.95 / 1000) = 0.04295 L 1L = 1000 mL 7. Calculating R PV = nRT R = PVn T = 0.9798 atm x 0.04295 L1.6786x10-3mol x 296.35 K = 0.084595 ~ 0.0846 atm L mol-1K-1 8. Percent error = experimental value of R - 0.08210.0821 x 100 = 0.0846 - 0.08210.0821 x 100 Percent error = 3.045 %