CaCO3(s) ⟶ CaO(s)+CO2(s)
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3,
therefore 0.0131 moles CO2 should also be formed.
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2
Therefore:
% Yield: 0.53/.576 x100= 92 percent yield
Answer : The percent yield for this reaction is, 91.9 %
Solution : Given,
Mass of carbon dioxide = 0.53 g
Mass of calcium carbonate = 1.31 g
Molar mass of carbon dioxide = 44 g/mole
Molar mass of calcium carbonate = 100 g/mole
First we have to calculate the moles of [tex]CaCO_3[/tex].
[tex]\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}=\frac{1.31g}{100g/mole}=0.0131moles[/tex]
Now we have to calculate the mass of carbon dioxide.
The balanced chemical reaction is,
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]CaCO_3[/tex] react to give 1 mole of [tex]CO_2[/tex]
So, 0.0131 mole of [tex]CaCO_3[/tex] react to give 0.0131 mole of [tex]CO_2[/tex]
Now we have to calculate the mass of carbon dioxide.
[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2[/tex]
[tex]\text{Mass of }CO_2=(0.0131mole)\times (44g/mole)=0.5764g[/tex]
Now we have to calculate the percent yield for this reaction.
[tex]\%\text{ yield of }CO_2=\frac{\text{Actual yield of }CO_2}{\text{Theoretical yield of }CO_2}\times 100=\frac{0.53g}{0.5764g}\times 100=91.9\%[/tex]
Therefore, the percent yield for this reaction is, 91.9 %
