Question

446 g of PbO, 46 g of NO2, and 16 g of O2, are
allowed to react according to the equation :-
PbO+2NO2+1/2O2 -: Pb(NO3) 2
The amount of Pb(NO3)2 that can be produced
is (At. wt. of Pb = 207) :-
(1) 331 g
(2) 662 g
(3) 165.5 g
(4) 132.5 g​

Answer 1

Answer:

662g

Explanation:

mole = $\frac{mass}{Mr\\}$, Mr stands for relative formula mass

PbO >>

mole = $\frac{446}{223}$  [ Mr of PbO = 207 + 16= 223] Ar( relative atomic mass)of                    

        = 2 mol                                                                   oxygen is = 16

ratio between PbO : Pb(NO3)2

                           1    :    1

the coefficients in the front of PbO and Pb(NO3)2 is both are 1.So, ratio is 1:1

Therefore if mole of PbO is 2 then the mole of Pb(NO3)2 is also 2 as both have the same ratio.

Pb(NO3)2 >>

Mass = Mole x Mr [ Mr of Pb(NO3)2 = 207 + (14 + 16 x 3) x 2 = 331]

        = 2 x 331                                                      Ar of nitrogen = 14

        = 662g