Question In Description

Question

3. A sample of aluminunm that weighed 0.050 g reãcted with excess 6.0 M HCI. Ả volume of

48.3 mL of H, gas was collected over water at 22.0°C and at an atmospheric pressure of 751 mm

Hg. Calculate the percent purity of the aluminum in the sample. (P_H20 =19.83 mm Hg at 22.0°C.)

Answer 1

First, write a correctly balanced equation for the reaction:

2Al(s) + 6HCl(aq) ==> 2AlCl₃(aq) + 3H₂(g) .. balanced equation

To find the amount of Al originally present, we will use the moles of H2 produced, and the stoichiometry of the balanced equation to find moles of Al. From there, we'll convert to grams and finally to % purity.

Ideal gas law: PV = nRT

P = pressure = 751 mm Hg - 19.83 mm Hg = 731.17 mm Hg (you must subtract vapor pressure of H2O)

V = volume in L = 48.3 ml x 1 L / 1000 ml = 0.0483 L

n = moles of H2 gas = ?

R = gas constant = 62.36 Lmm Hg / Kmol

T = temperature in Kelvin = 22ºC + 273 = 295 K

Solving for moles of H2:

n = PV/RT = (731.17)(0.0483) / (62.36)(295)

n = 1.92x10^-3 moles H2

moles of Al originally present = 1.92x10^-3 moles H2 x 2 mol Al / 3 mol H2 = 1.28x10^-3 mols Al

mass of Al originally present = 1.28x10^-3 mols Al x 26.98 g / mol = 0.0345 g Al

% purity = 0.0345 g / 0.050 g (x100%) = 69.1% pure

Answer 2

There are a few parts to this question. I listed out what you do for each part but did not include my work. If you have any questions about a particular part please feel free to reach out.

The first thing I did was convert mmHg to kiloPascals. I am used to using a value of 8.314 for R in the ideal gas law (PV=nRT), so I needed to have pressure in kPa. If you used mmHg you will have to use a different value for R (you can prob just google it).

1. You must write out an equation for the reaction. You must know that Al typically will give up 3 electrons which means it must form an ionic compound with 3 chlorines. Then you just balance out the equation with the correct numbers. These numbers will be used in part 3.
2. You can use the ideal gas law (PV=nRT) to figure out how many moles of Hydrogen gas was produced. Solving for n will give you moles of H2 gas.
3. Your answer from part 2, and the equation from part 1 are used to figure out how much Aluminum must have been present at the start of the reaction.
4. Once you know how much aluminum must have been in the sample, you can put it over the mass of the sample and multiply by 100 to get your precent purity.

I got a final answer of approximately 69% pure. Hope this helps. I would be happy to explain myself farther or write out my work for a particular part if anything is confusing.

Hope this helps :)