Lead (II) Oxide Reaction?

Please help, I don't know what to do :)

What mass of Lead (II) Oxide is obtained by heating 33.1g of Lead (II) Nitrate in the reaction:

2Pb(NO3)2 --> 2PbO + 4NO2 + O2

Any help much appreciated. Thank you.

Answers:
2 moles of the nitrate make 2 moles of the oxide.
Or 1 mole of itrate makes one mole of oxide
or 331 grms of nitrate make 223 grms of oxide
So 33.1 makes 22.3 grms.
The answers above contain errors.
Since Pb(NO3)2 molecule weight is 331 and PbO's molecule weight is 223, 22.3g of Lead Oxide will be obtained.

Reference: atomic table
Pb =207
N=14
O=16
n(Pb(NO3)2)= 331/33.1 = 10 moles

ratio 2 to 2

10 moles of PbO is required

m = n x Mw = 10 x 223.2 = 2232g of PbO will be obtained.

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