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446 g of PbO , 46 g of NO(2) and 16 g of...

446 g of PbO , 46 g of `NO_(2)` and 16 g of `O_(2)` are allowed to react according to the equation :-
`PbO+2NO_(2)+(1)/(2)O_(2)toPb(NO_(3))_(2)`
The amount of `Pb(NO_(3))_(2)` that can be produced is (At . Wt. of Pb =207) :-

A

331 g

B

662 g

C

165.5 g

D

none of these

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