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What percent by mass of lead nitrate [Pb...

What percent by mass of lead nitrate `[Pb(NO_(3))_(2)]` is reduce when heated strongly?

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`underset(2(207+2xx14+6xx16)=662g)(2Pb(NO_(3))_(2)) to underset(2(207+16)=446g)(2PbO)+4NO_(2) uarr+O_(2)uarr`
On strong heating `Pb(NO_(3))_(2)` loses its mass as `NO_(2)` and `O_(2)` escape out as gases and only PbO is left behind as solid residue (yellow cloured).
the above equation shows that 446g of PbO is left behind as residue when 662g of `Pb(NO_(3))_(2)` is strongly heated.
`therefore` Loss in mass`=(662-446)g=216g`
`therefore`Percentage of loss by mass`=(216)/(662)xx100=32.62`
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