Question

Experiment 12. Generating Hydrogen Cas A. Percent yield of hydrogen gas Mass of magnesium -0.051 Volume of gas (actual) Temperature of H gas Atmospheric pressure (see barometer) 20 °C 24 mmHg 26PmHg mmHg Vapor pressure of water there mmtig Partial pressure of He gas Calculate the moles of hydrogen gas that should have been produced, starting with the mass of Mg and using the balanced equation. Mg (8) + 2 HCI (aq) → MgCl(aq) + H2 (8) 0.08ling Imoling . Imol H2 -0.00333 labmoltz 24.319mg imong .00333 moles Hz Using the ideal gas law, calculate the theoretical yield of hydrogen gas. Ry=nRT DV - teatma nT (0.0821 live than 12938) - L H2 = mL H2 (theoretical yield) Calculate the percent yield. Percent yield: 125

Answer 1

The theoretical volume of H2 is calculated:

V = n * R * T / P = 0.00333 mol * 0.082 atm * L / mol * K * 293 K / 0.976 atm = 0.082 L = 82 mL

The yield percentage is calculated:

% yield = 10 mL * 100/82 mL = 12.20%

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