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Topic: Net ionic equation  (Read 1838 times)

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Offline EO_TheTruth

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Net ionic equation
« on: October 26, 2016, 06:57:39 PM »
Hello,
This is my first time finding the net ionic equation of a double replacement reaction. I just wanted to see if I did it correctly.

Offline EO_TheTruth

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Re: Net ionic equation
« Reply #1 on: October 26, 2016, 07:01:41 PM »
Double replacement reaction question

Given
Na2(CO3)Ca(OH)2 -->

Solved
Na2(CO3) + Ca(OH)2 --> Na(OH) + 2Ca(CO3)

Net Ionic Equation
CO3+Ca(OH)2----> OH+2Ca+2CO3

Spectator ions
Na---> Na

Offline AWK

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Re: Net ionic equation
« Reply #2 on: October 26, 2016, 07:06:24 PM »
a, b - wrong
Moreover correct formulas are:
Na2CO3 and CaCO3
AWK

Offline EO_TheTruth

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Re: Net ionic equation
« Reply #3 on: October 26, 2016, 07:08:57 PM »
What's wrong with a and b? Is c wrong to?

Offline EO_TheTruth

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Re: Net ionic equation
« Reply #4 on: October 26, 2016, 07:13:40 PM »
a, b - wrong
Moreover correct formulas are:
Na2CO3 and CaCO3

What are you referring to? By
Moreover correct formulas are:
Na2CO3 and CaCO3

Offline AWK

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Re: Net ionic equation
« Reply #5 on: October 26, 2016, 07:18:10 PM »
You write down Na2(CO3) and Ca(CO3) - this make some difference.

c- almost correct - spectator ion is on both side of equation then Na+ is sufficent
a- wrong coefficient
b- anion with + charge?, Ca(OH)2 in solution is ionic, CaCO3 is insoluble in water then we write it as if  non-ionic.
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Offline AWK

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Re: Net ionic equation
« Reply #6 on: October 26, 2016, 07:24:23 PM »
NaOH should be also without brackets
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