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Topic: Mixture Problem (yes, another one)  (Read 2791 times)

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Offline sega18

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Mixture Problem (yes, another one)
« on: January 30, 2012, 10:11:57 PM »
Mixture of CaO and CaCO3 weighing 3.592g.

CaCO3 -> CaO + CO2

Mass of CaO left in crucible: 2.270g

Find % comp.

x = mass CaO in mixture
y = mass CaCO3
z = mass CaO formed from decomposition

x + y = 3.592
x + z = 2.270

z = .82653y (y g CaCO3 X 1 mol/68.08 g CaCO3 = z g CaO X 1 mol/56.277g)

x + y = 3.592
x + .82653y = 2.270

by graphing calculator:
x = -4.0289
y = 7.6209

What did I do wrong? This is bugging the s#*$ out of me.
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Offline UG

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Re: Mixture Problem (yes, another one)
« Reply #1 on: January 31, 2012, 12:54:04 AM »
May I suggest an alternative approach where you let x = moles of CaO in the mixture and y = moles of CaCO3 in the mixture. Can you write two simultaneous equations with these two variables?
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Offline Borek

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Re: Mixture Problem (yes, another one)
« Reply #2 on: January 31, 2012, 05:14:17 AM »
Your approach is correct, your molar mass of CaCO3 is not.
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Offline sega18

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Re: Mixture Problem (yes, another one)
« Reply #3 on: January 31, 2012, 09:31:30 AM »
What a dumb error. I hate myself.
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