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 100 mL 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 0.0449 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)

The final volume was 32.34ml and the inital volume was 1.02 ml. Actual volume is 32.34ml - 1.02ml = 31.32ml

Mg(s) = 0.0449g

Questions (show work):

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

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

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

Answer 1

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

number of moles of HCl = 5 * 10^-3 * 3 = 0.015 moles

number of moles of Mg = 0.0449g/24g/mole = 0.001871 moles

Actual volume of H2 gas is 32.34ml - 1.02ml = 31.32ml

1 mole of Mg reacts with 2 moles of HCl

0.00187 moles of Mg reacts with 2 * 0.00187 moles of H2 gas

                                = 0.00374

so Mg is the limiting reagent

1 mole of Mg produces 1 mole of H2 gas

0.00187 moles of Mg produces 1 *0.00187 moles of H2 theoritically

number of moles of H2 produce theoritically = 0.00187 moles

number of moles of H2 gas produced in experiment n = PV/RT

                                                 n = 1 * 31.32/(0.0821 * 298)
                                            
number of moles of H2 produced in experiment n = 0.00128 moles

percent of yoeld = Actual moles of H2(g)/expected moles of H2(g)×100%

percent yield = 0.00128/0.00187 * 100 = 68.45 %