Question

Experiment 12: Generating Hydrogen Gas Data: A. Percent yield of hydrogen gas Mass of magnesium 0.04298 Volume of H2 gas (actual) 42.ml 0.0426 24.0°C 297 Temperature of H, gas Atmospheric pressure (see barometer) 765.8 mmHg a2.4 mmHg Vapor pressure of water Partial pressure of Hą gas 743.4 mmHg 0.976 1 atm Calculate the moles of hydrogen gas that should have been produced, starting with the mass of Mg and using the balanced equation. mg = 24.31 Mg(s) + 2 HCI (aq) → MgCl(aq) + H: (g) mass mg moreng moet moles H; Using the ideal gas law, calculate the theoretical yield of hydrogen gas. LH- mL H(theoretical yield) Calculate the percent yield. Percent yield:

Answer 1

мя (8) + 2H4 сая, 1) hя а, (4) +н, (4) molas man 1 м = 24 -30s 3. molg ng = given man molar mass | mol 1 м2 = 0.00 111mo - ооч-% 2Ч. 30s a mol:

mol og mg = 0.00177 moll 1 mol of mg produced Imol H₂ 0.00177 mol g mg produced 0.00172 mol H2 12 obtained = 0.00177 mol PV = nRT P = Pressure V = Volume n= mol R= gas constant T- Temp. I atm = 760 mmHg p= 0.9781 atm V = nRT = 0.00177md X 0.0821 Lamm x 297K mork 0.9781 atox = 0.04412 volume = 0.04412 = 44.1mL ( 1L = 1000 ml) Thoeretical volume = 44.1 mL Experimente = 42.6 m2 l yield= 4206 ml Neo= 96.4.1- 44.1 me