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Chemistry 20 Online
OpenStudy (anonymous):

Can somebody please please please help me with balancing equations? :( i'm so sad i can't figure it out.

OpenStudy (espex):

What question?

OpenStudy (anonymous):

lead (II) nitrate --> lead (II) oxide + nitrogen dioxide + oxygen

OpenStudy (anonymous):

i just don't get it :/

OpenStudy (espex):

First write out the empirical formula so you can balance it.

OpenStudy (espex):

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

OpenStudy (anonymous):

Okay, but I don't understand how to get it to the ratio.

OpenStudy (anonymous):

notice there is an odd number of O on the right and always an even number of the left. Therefore, you must have 2 PbO

OpenStudy (espex):

The ratios IS the balancing portion. You will deduce them when you put in the right amount of each.

OpenStudy (anonymous):

Well, yes. But it's just confusing like how to balance them.

OpenStudy (anonymous):

start with one element

OpenStudy (anonymous):

then go element by element

OpenStudy (anonymous):

Pb(NO3)2 -> 2 PbO + NO2 + O2 you need 2 Pb on the left to balance the 2 Pb on the right 2Pb(NO3)2 -> 2 PbO + NO2 + O2

OpenStudy (anonymous):

now you need 4 N on the right to balance the 4 N on the left 2Pb(NO3)2 -> 2 PbO + 4NO2 + O2

OpenStudy (anonymous):

there's 12 O on the left, and 12 on the right, so it's balanced

OpenStudy (anonymous):

Oh okay thank you so much!

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