# How do you balance this equation?

Ce2(SO4)3 + HgSO4 + H2SO4 + HNO3 + CO2 = Ce(SO4)2 + Hg2(CN)2 + H2O

I don't just want the answer, I want to know how to get the answer

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Ce2(SO4)3 + HgSO4 + H2SO4 + HNO3 + CO2 = Ce(SO4)2 + Hg2(CN)2 + H2O

examine the ones which change oxidation number

Ce+3 → Ce+4 + 1e-

Hg+2 +1e- → Hg+1

N+5 + 8e- → N-3

C+4 +2e- → C+2

but these last three are all together in Hg2(CN)2

there are two of each so collectively they gain 22e-

let's take care of the 'two of each' thing first

Ce2(SO4)3 + 2 HgSO4 + H2SO4 + 2 HNO3 + 2 CO2 = Ce(SO4)2 + Hg2(CN)2 + H2O

now each Ce loses 1 e- so we need 22 on each side

11 Ce2(SO4)3 + 2 HgSO4 + H2SO4 + 2 HNO3 + 2 CO2 = 22 Ce(SO4)2 + Hg2(CN)2 + H2O

OK, the electrons are balanced

now let's do the sulfate ions

there are 44 on the right and we only want to change H2SO4 so we don't unbalance electrons

11(3) + 2 + x = 44 so x is 9

11 Ce2(SO4)3 + 2 HgSO4 + 9 H2SO4 + 2 HNO3 + 2 CO2 = 22 Ce(SO4)2 + Hg2(CN)2 + H2O

now the H's

there are 18+2 = 20 on the left so we need 10 H2O on the right

11 Ce2(SO4)3 + 2 HgSO4 + 9 H2SO4 + 2 HNO3 + 2 CO2 = 22 Ce(SO4)2 + Hg2(CN)2 + 10 H2O

the only thing left to check are the O's

we can make that easier by checking only the non-sulfate O's because sulfates are balanced already

There are 6 + 4 = 10 on the left in 2HNO3 and 2CO2

and there are 10 on the right in 10H2O

so it's all balanced

11 Ce2(SO4)3 + 2 HgSO4 + 9 H2SO4 + 2 HNO3 + 2 CO2 → 22 Ce(SO4)2 + Hg2(CN)2 + 10 H2O

a pain in the neck, but doable

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• I won't give ya the answer ill try and let ya work it out by ya self but what you need to know is that in a chemical reaction no matter is lost or gained just chaged witch means anything on the left will be on the right just in a diffrent form and all you need to do is write down all the simbols down from each side and count them so you know how many is on each side then to balance it put numbers in front of the simbol to change the amount of it do not change the number after beacsue that changes the symbol to a diffrnt thing

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• Old Science Guy has the correct answer.

Here are seven more problems like yours:

http://chemteam.info/Redox/Redox-FourEquations.htm...

Each of then required four half-reactions in order to balance.

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