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### Topic: Problem of the week - 30/01/2012  (Read 16640 times)

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#### Borek ##### Problem of the week - 30/01/2012
« on: January 30, 2012, 09:22:47 AM »
When 0.5243g piece of lithium was thrown into a watery liquid, it reacted violently, giving off a colorless gas. The gas was collected over water in an inverted graduated cylinder. Volume of the gas - at the temperature of 28ºC and atmospheric pressure of 745 mmHg - was found to be 990 mL. In a separate experiment density of the gas - as transferred from the cylinder, at the same temperature and pressure - was measured as 0.1809 g/L. Remaining solution was saturated with carbon dioxide and dried out, leaving white solid. Mass of the solid was found to be 2.790 g.

Can you name the liquid?
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#### Borek ##### Re: Problem of the week - 30/01/2012
« Reply #1 on: January 31, 2012, 07:13:08 AM »
No takers?

Too easy? Too difficult? You get numbers you don't understand?
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#### Ann1234

• Regular Member
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• Mole Snacks: +4/-0
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« Reply #2 on: January 31, 2012, 08:43:34 PM »
I got D20 (heavy water)---

this is what I did...First of all I calculated the moles of Li (0.07554 moles) with the mass and atomic mass of Li.

"When 0.5243g piece of lithium was thrown into a watery liquid,"

Li + ?      ====>   ? + ?
0.07554

"The gas was collected over water in an inverted graduated cylinder. Volume of the gas - at the temperature of 28ºC and atmospheric pressure of 745 mmHg - was found to be 990 mL. "

Then, using the information from the gas collected over water, I got the number of moles of the dry gas (0.0377 moles). So I divided 0.07554 by this number and that ratio is 2, meaning that the ratio Li/gas in the equation is 2:1

"In a separate experiment density of the gas - as transferred from the cylinder, at the same temperature and pressure - was measured as 0.1809 g/L. "

Then, with the density of the gas, I have some trouble at first because I neglected the water vapor that was in the cylinder. I got the molar mass from there and it didn't make sense to me so I realized that I needed to include the water since the gas that was transferred from the cylinder had water vapor mixed with it

So I got the water vapor pressure from a table and at 28C: 28.3 torr. With this value, plus T=301 K, plus v= 0.99 L, I got the moles and then the mass of water that was inside the cylinder:

mass H2O = 0.027 g

From the density of the gas, I got the mass of gas in 0.990 L: 0.1809 g/L X 0.99L = 0.179 grams

then from this mass I subtracted mass of water vapor, to get the mass of my gas (dry)

mass gas mixture = 0.179 g = mass dry gas + 0.027 grams

therefore: mass gas (dry) = 0.152 grams

With this mass and the number of moles from the beginning (0.0377) I got the molar mass of the gas

M = mass/moles = 4.027 grams/mole

With this low molar mass, the first thing that came out through my mind was H2, and the reaction that gives H2 as a product that I know is:

2Li + 2H20 = 2LiOH + H2

It made sense that the ratio Li: H2 is 2:1. But H2 molar mass is 2g/mol, so I though that this gas could be a isotope of Hydrogen, and I thought on deuterium since its molar mass is 4.029 g/mol.

So far my reaction would be:

2Li + 2D20 = 2LiOD + D2

Now, regarding the last part of the experiment:

"Remaining solution was saturated with carbon dioxide and dried out, leaving white solid. Mass of the solid was found to be 2.790 g."

Now, assuming that the reaction occurs between LiOD and CO2, I wrote the products and balanced the equation:

2LiOD + CO2 --> Li2CO3 + D2O
2.790 g

Then I got the moles of lithium carbonate (0.03775 moles) and from there I knew that moles LiOD = 0.07554 moles since the ratio is 2:1. I confirmed that moles Li = moles LiOD and it make sense that lithium carbonate is the product since after drying it will come out as a white solid.

I hope I got this! Thank you for such challenging problem. =) (And excuse me for my English (I am not native speaker))
« Last Edit: January 31, 2012, 09:43:16 PM by Ann1234 »

#### XGen

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• Mole Snacks: +9/-4 ##### Re: Problem of the week - 30/01/2012
« Reply #3 on: January 31, 2012, 09:17:07 PM »

#### Borek ##### Re: Problem of the week - 30/01/2012
« Reply #4 on: February 01, 2012, 05:24:19 AM »
I got D20 (heavy water)

You got it right. And you nicely navigated around the vapor pressure thing - I was tempted to list 28.3 mmHg at 28ºC in the question, but then it would be too obvious why the calculated molar mass of the gas takes a rather random value.