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Topic: Decomposition of Calcium Carbonate and Magnesium Carbonate  (Read 8311 times)

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Offline xstrae

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Decomposition of Calcium Carbonate and Magnesium Carbonate
« on: October 01, 2006, 02:09:53 AM »
A sample of CaCO3 and MgCO3 weighing 192g is ignited to constant weight of 104g. What is the composition of the mixture? Also calculate the volume of CO2 evolved at NTP.

The problem seems simple enough. But I am not able to get the specified answer in the book for some reason. Since,the reduction in mass on heating is 88g, this must account for the CO2 evolved. So 2 moles of CO2 i.e, 44.8litres of CO2 are evolved. How do I find the composition?
« Last Edit: October 02, 2006, 02:58:26 PM by geodome »
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Offline Will

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Re: Stoichiometry Problem
« Reply #1 on: October 01, 2006, 02:46:44 AM »
100.086x + 84.313(2-x) = 192   (2-x because you know that there are 2 moles of MgCO3 and CaCO3)
solve for x:
x = 1.481899
therefore
74.09% is CaCO3
25.91% is MgCO3
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Offline Donaldson Tan

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Re: Stoichiometry Problem
« Reply #2 on: October 01, 2006, 02:47:20 AM »
let Mi be molar mass of component i (unit: g/mol)

let x be no. of moles of CaCO3
let y be no. of moles of MgCO3

x.MCaCO3 + y.MMgCO3 = 192g

CaCO3 -> CaO + CO2
MgCO3 -> MgO + CO2

x.MCaO + y.MMgO = 104g

You have 2 variables and 2 simultaneous equations - solve for x and y.
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Offline xstrae

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Re: Stoichiometry Problem
« Reply #3 on: October 01, 2006, 06:20:28 AM »
ok thanks geodome. I got 44.4g of MgCO3. My book says 42g but I suppose its good enough. :)
Will, it was not mentioned that only 1 mole of each is present. Anyway thanks for the *delete me*
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Offline Will

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Re: Stoichiometry Problem
« Reply #4 on: October 01, 2006, 11:47:33 AM »
Quote from: konichiwa2x on October 01, 2006, 06:20:28 AM
Will, it was not mentioned that only 1 mole of each is present. Anyway thanks for the *delete me*

Sorry, I realise what I said wasn't very clear. I just meant that you know two moles of CO2 would be produced, so two moles of (CaCO3+MgCO3) would have been present, i.e. the number of moles of each carbonate added together would equal two. That is why you can write it like this: 100.086x + 84.313(2-x) = 192. It doesn't matter which one is (2-x) and which one is x, as long as you know what it corresponds to. If you wanted to be more accurate you could use 1.99959 instead of 2.

I got the mass of MgCO3 to equal 43.68g, which is also quite far off 42g.
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Offline Donaldson Tan

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Re: Stoichiometry Problem
« Reply #5 on: October 01, 2006, 06:41:13 PM »
Quote from: konichiwa2x on October 01, 2006, 06:20:28 AM
My book says 42g but I suppose its good enough. :)

I got exactly 42g.

Quote from: geodome on October 01, 2006, 02:47:20 AM
x.MCaCO3 + y.MMgCO3 = 192g
x.MCaO + y.MMgO = 104g

x = 3/2; y = 1/2

mass of MgCO3 = y.MMgCO3 = (1/2)*84 = 42g
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"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline xstrae

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Re: Stoichiometry Problem
« Reply #6 on: October 02, 2006, 10:05:59 AM »
sorry for mistaking you Will. Anway, I understood it now. thank you guys!
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