Respuesta :
Answer: The percentage yield of [tex]CO_2[/tex] is 90.26%.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of [tex]CaCO_3[/tex] = 80 g
Molar mass of [tex]CaCO_3[/tex] = 100 g/mol
Plugging values in equation 1:
[tex]\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8 mol[/tex]
For the given chemical equation:
[tex]CaCO_3\rightarow CaO+CO_2[/tex]
By the stoichiometry of the reaction:
If 1 mole of [tex]CaCO_3[/tex] produces 1 mole of [tex]CO_2[/tex]
So, 0.8 moles of [tex]CaCO_3[/tex] will produce = [tex]\frac{1}{1}\times 0.8=0.8mol[/tex] of [tex]CO_2[/tex]
Molar mass of [tex]CO_2[/tex] = 44 g/mol
Plugging values in equation 1:
[tex]\text{Mass of }CO_2=(0.8mol\times 44g/mol)=35.2g[/tex]
The percent yield of a reaction is calculated by using an equation:
[tex]\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100[/tex] ......(2)
Given values:
Actual value of [tex]CO_2[/tex] = 35.2 g
Theoretical value of [tex]H_2CO_3[/tex] = 39 g
Plugging values in equation 2:
[tex]\% \text{yield of }CO_2=\frac{35.2g}{39g}\times 100\\\\\% \text{yield of }CO_2=90.26\%[/tex]
Hence, the percentage yield of [tex]CO_2[/tex] is 90.26%.
