Answer :
2Pb(NO₃)₂ = 2PbO + 4NO₂ + O₂
m{Pb(NO₃)₂}=9.9 g
m'{PbO}=5.5 g
M{Pb(NO₃)₂}=331.2 g/mol
M{PbO}=223.2 g/mol
w-?
w=m'{PbO}/m{PbO}
m{PbO}=M{PbO}m{Pb(NO₃)₂}/M{Pb(NO₃)₂}
w=m'{PbO}M{Pb(NO₃)₂}/[M{PbO}m{Pb(NO₃)₂}]
w=5.5*331.2/[223.2*9.9]=0.8244 (82.44%)
m{Pb(NO₃)₂}=9.9 g
m'{PbO}=5.5 g
M{Pb(NO₃)₂}=331.2 g/mol
M{PbO}=223.2 g/mol
w-?
w=m'{PbO}/m{PbO}
m{PbO}=M{PbO}m{Pb(NO₃)₂}/M{Pb(NO₃)₂}
w=m'{PbO}M{Pb(NO₃)₂}/[M{PbO}m{Pb(NO₃)₂}]
w=5.5*331.2/[223.2*9.9]=0.8244 (82.44%)
82% the per cent yield of the decomposition reaction.
What is decomposition reaction?
The chemical reaction in which the reactants breaks or splits to produce two or more products is called a decomposition reaction.
How to calculate the percentage yield?
The reaction is given as,
[tex]\rm 2Pb(NO_{3} )_{2}(s) \rightarrow2PbO(s) + 4NO_{2}(g) + O_{2} (g)[/tex]
Given,
- Mass [tex]\rm (m_{1})[/tex] of lead nitrate = 9.9 g
- Mass [tex]\rm (m_{2})[/tex] of lead oxide = 5.5 g
- The molar mass [tex]\rm (M_{1})[/tex]of lead nitrate = [tex]331.2 \;\rm g \;mol ^{-1}[/tex]
- The molar mass [tex]\rm (M_{2})[/tex] of lead oxide = [tex]223.2 \;\rm g \;mol^{-1}[/tex]
Moles can be calculated by,
[tex]\rm Moles (n) =\dfrac {\rm Mass (m)}{\;\rm Molar \;mass (M)}[/tex]
The percentage yield can be given as,
[tex]\rm Percentage \;yield = \dfrac{\rm Moles\; of \;lead \;oxide}{ Moles \;lead \;nitrate} \times 100\%[/tex]
Substituting values we get,
[tex]\begin{aligned}\% \rm yield &=\dfrac{ \dfrac{5.5\;\rm g}{223.2 \;\rm g \;\rm mol ^{-1}}}{\dfrac{9.9 \;\rm g}{331.2\rm \; g \;\rm mol^{-1}}} \times 100\\\\&= \dfrac{5.5 \;\rm g \times 331.2\;\rm g\;mol^{-1}}{223.2\;\rm g\;mol^{-1} \times 9.9\;\rm g} \times 100\\\\&=82 \%\end{aligned}[/tex]
Therefore, 82% is the yield of the decomposition reaction.
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