Question

Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if $9.9g$ of $Pb{\left( {N{O_3}} \right)_2}$ are heated to give $5.5g$ of $PbO$ ?

Answer 1

Lead nitrate is an inorganic compound with molecular formula of $Pb{\left( {N{O_3}} \right)_2}$ . It can be decomposed into lead oxide, nitrogen dioxide and oxygen gas when heated. The percentage yield of reaction can be determined by dividing the moles of lead oxide with the moles of lead nitrate.
Formula used:
$\% yield = \dfrac{{{n_{PbO}}}}{{{n_{Pb{{\left( {N{O_3}} \right)}_2}}}}}$
${n_{PbO}}$ is number of moles of lead oxide
${n_{Pb{{\left( {N{O_3}} \right)}_2}}}$ is number of moles of lead nitrate
Complete step by step solution:
Inorganic compounds are the compounds that were formed by the combination of atoms other than carbon and hydrogen. Lead nitrate is a compound that was formed by the combination of lead, nitrogen and oxygen atoms. The molecular formula of lead nitrate is $Pb{\left( {N{O_3}} \right)_2}$ .
Lead nitrate on heating undergoes a decomposition reaction to produce lead oxide, nitrogen dioxide and oxygen gas.
$Pb{\left( {N{O_3}} \right)_2}\xrightarrow{\Delta }PbO + 2N{O_2} + \dfrac{1}{2}{O_2}$
Given mass of lead nitrate is $9.9g$ and the molar mass is $331.2gmo{l^{ - 1}}$
Given mass of lead oxide is $5.5g$ and the molar mass is $223.2gmo{l^{ - 1}}$
Substitute these values in the above formula, as the ratio of mass and molar mass gives the moles.
$\% yield = \dfrac{{\dfrac{{5.5}}{{223.2}}}}{{\dfrac{{9.9}}{{331.2}}}} \times 100 \approx 80\%$
Thus, the percent yield of the decomposition reaction is $80\%$

Note:
Generally, inorganic compounds on heating forms oxides along with the liberation of oxygen gas. The percentage yield can be calculated from the balanced stoichiometric coefficients equation, in the above reaction there is $1:1$ mole ratio of lead oxide and lead nitrate. Thus, the ratio for the percentage yield was calculated for $1:1$ mole ratio.