Question

6.5 g of lead combines directly with 1.0 g of oxygen to form lead peroxide. Lead peroxide is also obtained by heating lead nitrate. Lead peroxide thus formed is found to contain 13.38% of oxygen. Show that the data illustrates the law of definite proportion.​

Answer 1

Answer:

In first reaction :

$\sf Pb + O_2 \longrightarrow PbO_2$

• Lead (Pb) = 6.5g
• Oxygen (O2) = 1.0 g

By law of conservation of mass, mass of lead peroxide found in this reaction is :

=> 6.5 + 1.0 = 7.5 g

% Composition of Oxygen in PbO2:

$\footnotesize\implies \% \: of \: oxygen \: in \: PbO_2 = \dfrac{mass \: of \: O_2 \: in \: compund}{Total \: mass \: of \: compound }\times 100$

$\footnotesize\implies \% \: of \: oxygen \: in \: PbO_2 = \dfrac{1.0}{7.5 }\times 100$

$\footnotesize\implies \% \: of \: oxygen \: in \: PbO_2 = 0.133\times 100$

$\footnotesize\implies \bf\% \: of \: oxygen \: in \: PbO_2 = 13.3 \: \%$

Second reaction:

$\sf Pb(NO_3)_2 \longrightarrow PbO_2$

% composition of oxygen in this reaction is 13.38%

• Percentage composition of oxygen in both the reactions are same that's why they follow the law of definite proportion.