Answer: 86.7 %
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of [tex]CaCO_3[/tex]
[tex]\text{Number of moles}=\frac{30.7g}{100g/mol}=0.307moles[/tex]
[tex]CaCO_3\rightarrow CaO+CO_2[/tex]
According to stoichiometry :
1 moles of [tex]CaCO_3[/tex] give = 1 mole of [tex]CO_2[/tex]
Thus 0.307 moles [tex]CaCO_3[/tex] give =[tex]\frac{1}{1}\times 0.307=0.307moles[/tex] of [tex]CO_2[/tex]
Mass of [tex]CO_2=moles\times {\text {Molar mass}}=0.307moles\times 44g/mol=13.5g[/tex]
[tex]\%\text{ yield }=\frac{\text{Actual yield}}{\text{Theoretical yield }}\times 100=\frac{11.7g}{13.5g}\times 100=86.7\%[/tex]
Therefore, the percent yield for the reaction is, 86.7%
