Balancing Equation help

Pb(NO3)2 --> PbO + NO2 + O2

really confused. I've tried putting 2 in front of NO2, but that would mean there would be 7 Oxygen on the product side.

Reply 1

TypicalNerd

Original post by Blackrose06

Pb(NO3)2 --> PbO + NO2 + O2

really confused. I've tried putting 2 in front of NO2, but that would mean there would be 7 Oxygen on the product side.

You can have fractions as coefficients if needed, then scale up the equation as a whole to get rid of the fraction afterwards.

Try finding a suitable fraction to put in front of the O2, so there are 6 oxygens on the RHS.

(edited 8 months ago)

Reply 2

Blackrose06

I'm still confused because I think i've messed up the lead part as well

Reply 3

TypicalNerd

Original post by Blackrose06

I'm still confused because I think i've messed up the lead part as well

In what way do you feel you’ve messed up the lead part? All the individual formulae are correct - it’s just the balancing that isn’t yet correct.

Let us start from the unbalanced equation:

Pb(NO3)2 —> PbO + NO2 + O2

Let us assume that there is just one Pb(NO3)2 on the LHS because the LHS is the simpler side of the equation, so it’s easier to work from. As such, you would only expect one Pb and two NO2’s to balance out the leads and nitrogens. Let’s therefore assume we have ‘a’ O2 molecules:

Pb(NO3)2 —> PbO + 2NO2 + aO2

Can you now write down an expression, in terms of ‘a’ for the number of oxygens on the right hand side?

(edited 8 months ago)

Reply 4

Blackrose06

Hang on, I may have got it, but I don't know if i've done it correctly:
2Pb(NO3)2 ---> 2PbO + 4NO2 +2O2

Reply 5

TypicalNerd

Original post by Blackrose06

Hang on, I may have got it, but I don't know if i've done it correctly:
2Pb(NO3)2 ---> 2PbO + 4NO2 +2O2

Not quite. Count the numbers of oxygens on each side of the equation.

Reply 6

Blackrose06

12 on the LHS, 14 on the RHS
So i wouldn't need to put a 2 in front of O2?

Reply 7

TypicalNerd

Original post by Blackrose06

12 on the LHS, 14 on the RHS
So i wouldn't need to put a 2 in front of O2?

Correct

Reply 8

Blackrose06

OK thanks.
I don't like balancing equations, it's so stressful for me, but it's also important

Reply 9

TypicalNerd

The way I was going about this originally was as follows:

Let the equation be

Pb(NO3)2 —> PbO + 2NO2 + aO2

In which case, the total number of oxygens on the RHS is given by the number of oxygens in 1 PbO + number of oxygens in 2 NO2’s + number of oxygens in a O2’s:

= (1 x 1) + (2 x 2) + (a x 2)
= 5 + 2a

Since there are 6 oxygens on the LHS and the numbers of oxygens on both sides must be the same, 5 + 2a = 6

This leads to a = 0.5, so:

Pb(NO3)2 —> PbO + 2NO2 + 0.5O2

Which can be doubled to get rid of the 0.5:

2Pb(NO3)2 —> 2PbO + 4NO2 + O2

Reply 10

Blackrose06

That makes sense. I didn't end up putting 0.5, but i can see what you did.
It was probably easier to do it that way
Thank you

Reply 11

TypicalNerd

Original post by Blackrose06

That makes sense. I didn't end up putting 0.5, but i can see what you did.
It was probably easier to do it that way
Thank you

You are most welcome.

I agree it’s ugly to use fractions and decimals, but when it isn’t immediately obvious that you need more than one of a particular compound on one side of the equation, sometimes it’s necessary to use them.