Quick Stoich & Gasses
The equation of the reaction provides you with the composition of the gas liberated. 

You assume ideal gas, which means that the gas particles are indistinguishable. This means the reaction equation could look like 
$$ \ce{2Pb(NO3)2 -&gt; 5X + 4PbO} $$
where 5 moles of gas are liberated for every 2 moles of lead nitrate.

Reforming the ideal gas equation to $n=pV/(RT)$ will lead you to the numerical value. The amount of lead nitrate will then be $$n_{\ce{Pb(NO3)2}} = \frac{2n}{5}$$

From that calculating the mass is a trivial task.