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Posted

I need help in my assignment to balance a chemical equation. Help me to solve it.

 

 

This is the equation.

 

Hg(OH)2 + H3Po4 = Hg3(Po4)2 + HOH

 

How to balance it?

 

Posted (edited)

Now, balancing O

3Hg(OH)2 + H3Po4 = Hg3(Po4)2 + 6HOH

 

Oh my god, got stucked.

 

Not O, but whole PO4

On right there is (PO4)2

So place 2 in front of H3PO4

 

and we're done

3Hg(OH)2+2H3PO4->Hg3(PO4)2+6H2O

 

ps. Po is Polonium..

Edited by Sensei
  • 1 month later...
Posted

I know that this was posted a while ago and it was sorted. But I prefer a more mathematical way of balancing the equation, as this is easier to generalise to any equation (and reduces the amount of jumping from the left and right hand side to make sure everything balances out). Essentially what you need to do is make sure that the number of each atom or group of atoms is the same on the right and left hand side.

 

So for all of the components of the original equation you can have an unknown in front of the component (this is what you use as the sociometric coefficient and balance the equation). This means that we can rewrite the equation as:

 

aHg(OH)2 + bH3Po4 -> cHg3(Po4)2 + dH OH

Now you know that the number of Hg atoms has to be the same on the left and right hand side this means that a=3c to make this true. The 3 comes from the fact that Hg3(Po4)2 has three Hg atoms in the molecule.

 

If we do that for all of the groups in the equation (Hg, OH, H, Po4) then we get the follow equations:

 

Hg: a=3c (1)

OH: 2a=d (2)

H: 3b=d (3)

Po4: b=2c (4)

 

What you can do now is set one of the values to 1 (in this case c is good to set to 1).

So if c=1:

a = 3 (from (1))

b = 2 (from (4))

d = 6 (from (3) and b=2)

 

Now that we've got all of the stoichiometric coefficients we can put this back into the original equation to get the balanced formula.

3Hg(OH)2 + 2H3Po4 -> 1Hg3(Po4)2 + 6H OH

 

If any of the coefficients are fractions then you need to multiply ALL of the coefficients by the value of the fraction on the bottom.

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