Question

Given the following experiment and results answer the following questions:

Add 5 mL of 3 M HCl in a test tube and piece together the apparatus shown in the picture below. The test tube should be closed with a rubber stopper that has a glass tube connected to rubber tubing inserted. Next, fill a container with half full with water. Fill a graduated cylinder to the brim with water and invert it into the container of water without creating any air pockets. Clamp the graduated cylinder in place and adjust the rubber tubing until the end is tucked inside the graduated cylinder. Weigh about 0.1 g of Mg ribbon. Open the test tube, quickly transfer the Mg into it and seal it with the rubber stopper. Watch for the gas formation and record the amount of gas formed. From the volume of gas calculate the molar volume of the hydrogen gas.    Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)         

Initial volume: 100 ml

Final volume: 78 ml   

Questions, Please show all work.

1.) From the volume of gas calculate the molar volume of the hydrogen gas.

2.) Calculate the amount of hydrogen gas expected from the reaction of HCl with Mg. Use stoichiometry, appropriate units and consider the significant figures.

3.)  Calculate the actual amount of hydrogen gas produced using the collected volume of gas.

4.) Calculate the % yield of the reaction. [Actual mass of H₂(g)/expected mass of H₂(g)]×100%

Answer 1

2)

Mg + 2H+ -------> Mg2+ + H2

moles of HCl or H+ = molarity * volume = 3 * 5*10^-3  = 0.015 mol

stoichiometric moles of H+ = 0.015 / 2 = 0.0075 mol

moles of Mg = 0.1 / 24.3 = 0.0041 mol

stoichiometric moles of Mg = 0.0041 / 1 = 0.0041 mol

since the moles of Mg are less so it is a limiting reagent

moles of H2 will produced = 0.0041 mol

PV = nRT

V = nRT / P = 0.0041 * 8.314 * 298.15 / 101,325 = 1.00*10^-4 m^3

V = 1.00*10^-4 * 10^3 L

V = 0.100 L

V = 100 mL

so amount of hydrogen gas expected from the reaction = 100 mL

1)

so the molar volume of hydrogen gas = 100 / 0.0041 = 24,390 mL = 24.39 L

3)

actual gas produced = 100 - 78 = 22 mL

4)

% Yield = [Actual mass of H₂(g)/expected mass of H₂(g)]×100%

= (22 / 100 ) * 100

% yield = 22 %