12.976g of lead combines with 2.004g of oxygen to form PbO2. PbO2 can also be produced by heating lead nitrate and it was found that the pecentage of oxygen present in PbO2 is 13.38% with the help of given information illustrate the law of constant composition.

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Solution

$S t e p I - C a l c u l a t i o n o f m a s s o f o x y g e n p r e s e n t$
$C a s e 1 = M a s s o f l e a d p e r o x i d e f o r m e d = 12.976 + 2.004 g r a m s$
$P b + O_{2} ⟶ {P b O}_{2}$

$H e n c e w e m a y s a y t h a t 14.980 g r a m s o f l e a d p e r o x i d e c o n t a i n s 2.004 g r a m o f o x y g e n$
$H e n c e 100 g r a m o f l e a d p e r o x i d e w i l l c o n t a i n = \frac{2.004}{14.980} \times 100 = 13.38 %$
$I n s e c o n d c a s e$
$L e a d n i t r a t e Δ \to L e a d o x i d e$
$P e r c e n t a g e o f o x y g e n p r e s e n t = 13.38 %$

$S t e p I I - C o m p a r i s o n o f % o x y g e n i n b o t h e x p e r i m e n t s$
$% O x y g e n i n l e a d o x i d e i n f i r s t e x p e r i m e n t = 13.38 %$
$% O x y g e n p r e s e n t i n l e a d o x i d e i n s e c o n d e x p e r i m e n t = 13.38 %$
$S i n c e t h e % c o m p o s i t i o n o f o x y g e n i n b o t h t h e e x p e r i m e n t s i s s a m e, h e n c e i t r e p r e s e n t s l a w o f c o n s t a n t p r o p r t i o n .$

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Similar questions

A student while heating solid lead nitrate taken in a test-tube would observe :
(1) white residue of PbO2
(2) green residue of NO₂
(3) yellow residue of PbO
(4) brown residue of NO

Calculate the percentage composition of oxygen in lead nitrate [Pb(NO₃)₂].
(Pb = 207, N = 14, O = 16)

Q. Assertion :The reactions with

H C l

and

S O_{2}

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P b (I V)

P b (I I)

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i . e .,

reduction hence

P b O_{2}

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P b^{2 +} > P b^{4 +}

, hence the reaction is spontaneous in forward side.

Q. A lead storage battery has initially

200 g

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200 g

P b O_{2}

, plus excess

H_{2} S O_{4}

. How long could this cell deliver a current of

10 a m p

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S O_{2}

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(ii) reacting sodium sulphite with dilute

H_{2} S O_{4}

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