This week\'s experiment involves the determination of the molar mass of aluminum

Question

This week's experiment involves the determination of the molar mass of aluminum by measuring the amount of hydrogen gas liberated when a sample of aluminum is treated with excess hydrochloric acid.

In the reaction between aluminum and hydrochloric acid, how many moles of hydrogen gas would be liberated when 4.184E-2 grams of aluminum are treated with excess hydrochloric acid?

Moles of hydrogen gas liberated =  mol

The atmospheric pressure was measured to be 0.9949 atm and the room temperature was 22.00oC. At this temperature, the vapor pressure of water is 19.80 torr. What would be the volume of the number of moles of hydrogen just calculated under these conditions? Enter the answer in mL.

Volume of hydrogen gas =  mL

The syringe used in this experiment is calibrated to 60 mL. However, there is a 'dead volume' at the needle end of the syringe that is uncalibrated. This volume equals 1.20 mL. Bearing in mind that the syringe is upside down, what would the reading on the syringe be that corresponded to the volume you have just calculated?

Reading on the syringe =  mL

Suppose that you have completed this week's experiment and have determined that the molar mass of aluminum is 25.77 g mol-1. What is the percent error of your experimental result?

Percent error of the experimental result =  %

Explanation / Answer

1.

2 Al(s) + 6HCl(aq) ----------> 2AlCl3(aq) + 3H2(g)

moles of Al = mass of Al / molar mass of Al = 4.184 x 10^-2 g / 26.98 g/mol

moles of Al = 1.55 x 10^-3 mol

2 mol of Al produces 3 mol of H2

1 mol of Al produces 3/2 mol of H2

So,

1.55 x 10^-3 mol Al produces H2 = 3/2 * 1.55 x 10^-3 mol

Moles of H2 produced = 2.325 x 10^-3 mol

Vapour pressure of water = 19.80 torr = 0.026 atm

So,

P = Atmospheric pressure - V.P of water

P = 0.9949 atm - 0.026 atm

P = 0.9689 atm

Using ideal gas equation

PV = nRT

V = nRT / P

V = 2.235 x 10^-3 mol * 0.0821 L-atm/mol-K * (273+22) K / 0.9689 atm

V = 0.05587 L

V = 0.05587 * 1000 mL

V = 55.87 mL

Reading on the syringe = V - dead volume

Reading on the syringe = 55.87 mL - 1.2 mL = 54.67 mL

Percent error of experimental result = | (experimental mass * 100 / actual mass) - 100 |

Only magnitude is required that is why mod (| | ) is given the sign does not matter

Percent error of experimental result = (25.77 g/mol * 100 / 26.98 g/mol) - 100

Percent error of experimental result = 4.485 %

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