Question

A solid mixture of weight $5g$ consisting of lead nitrate and sodium nitrate was heated below ${600^o}C$ until the weight of the residue was constant. If the loss in weight is $28.0\%$, the weight of lead nitrate in the mixture is:
A. $3.32g$
B. $1.68g$
C. $3.87g$
D. $4.12g$

Answer 1

The number of moles of reactants and products should be balanced such that the loss in reactants shall compensate with the loss in the products. The two balanced chemical equations of the two compounds in the mixture shall be determined so that there shall not be any stoichiometric defaults.

Complete answer:
The additional loss in weights due to the formation of nitrogen dioxide and oxygen gases is
$= \dfrac{5}{{100}} \times 28 = 1.4g$
The chemical equations for the decomposition of lead nitrate and sodium nitrate are as follows:
$Pb{(N{O_3})_2} \to PbO + N{O_2} \uparrow + N{O_2} + {O_2}$
$\text{xg (x - y)g yg}$
$NaN{O_3} \to NaN{O_2} + {{dfrac{1}{2}}O_2}$
($5 - x$ ) ($3.6 - x + y$) ($1.4 - y$)
Where, $x =$ weight of lead nitrate taken initially
$y =$ weight of oxygen released
The number of moles of lead nitrate [$Pb{(N{O_3})_2}$ ] = number of moles of lead oxide [$PbO$ ]
As we know, the number of moles = $n = \dfrac{w}{{{M_w}}}$
Applying this in the above relation, we have:
$\dfrac{x}{{331}} = \dfrac{{(x - y)}}{{223}}$ …(i)
This also means that $331g$ of $Pb{(N{O_3})_2}$ gives $223g$ of$PbO$.
Similarly, The number of moles of sodium nitrate [$NaN{O_3}$ ] = number of moles of sodium nitrite [$NaN{O_2}$].
$\dfrac{{5 - x}}{{85}} = \dfrac{{3.6 - x + y}}{{69}}$ …. (ii)
As we can see that we have got two variables and two equations. SO equating both the equations and on solving, we have:
Mass of lead nitrate [$Pb{(N{O_3})_2}$] = $x = 3.324g$
Mass of sodium nitrate [$NaN{O_3}$] = $5 - x = 5 - 3.324 = 1.676g$

So, the correct answer is Option A.

Note:
Lead nitrate is used in the manufacture of matches and special explosives such as lead azide and pigments (such as in lead paints) for dyeing and printing calico and other textiles, and in the general manufacture of lead compounds such as lead oxide.