Question

i need help with the moles of H2 gas please

Trial 1 Trial 2 Mass of Mg ribbon (e) 10.0541 g 0.07839 Barometric pressure (mmHg) (Proom) 1 75 66 matts 756 mutta Temperature of the room (°C) 224 °C 22.4 °C Temperature of water inside beaker (°C) 23.0°C 22.8°C (T2) Pwater (mm Hg) Vapor pressure of water found in table 21.068 van Hg 20.615 mm Hg at the end of the experiment, based on the temperature of water inside the beaker. Volume from graduated cylinder (mL) 60.2mL 82.4 ml (V2) CALCULATIONS • Write the balanced chemical equation for the reaction between Mg and HCI. Mgr 2 HCl -> HgCl2 + H₂ . Calculate the number of moles of Mg that were used for each trial. 0.0541g Mg. Imol h 19. 24.005 -0.0022 mal 0.07839. Imal Mg. Sg. 24,3059 = 0.00 32 mol

Answer 1

No of moles of H2 gas is equal to the no of moles mg in each trail.

Since according to reaction what it says

Mg + 2HCl ---->> MgCl2 + H2

So 1mole of Mg gives you the 1mole of H2 gas.

So in trail 1 you got 0.0022mol of mg and also for H2 gas it will be 0.0022mol (1:1 mole ratio).

Similarly for trail 2 you got moles of mg =0.0032mol and also for H2 gas it will be 0.0032mol.