Answer: a) 74%
b) 59%
Explanation:
Atom economy gives how much desired product is obtained compared to amount of starting materials.
[tex]\text {Atom economy}=\frac{\text {molecular weight of desired product}}{\text {molecular weight of all products}}\times 100%[/tex]
For reaction a:
[tex]2CuO(s)+C(s)\rightarrow 2Cu(s)+CO_2(g)[/tex]
[tex]\text {atom economy of copper}=\frac{\text {molecular weight of desired product}}{\text {molecular weight of all products}}\times 100%[/tex]
[tex]\text {atom economy of copper}=\frac{2\times 63.5}{2\times 63.5+12+2\times 16}\times 100%=74\%[/tex]
For reaction b:
[tex]CuO(s)+CO(g)\rightarrow Cu(s)+CO_2(g)[/tex]
[tex]\text {atom economy of copper}=\frac{\text {molecular weight of desired product}}{\text {molecular weight of all products}}\times 100%[/tex]
[tex]\text {atom economy of copper}=\frac{63.5}{2\times 63.5+12+2\times 16}\times 100%=59\%[/tex]
Thus atom economy for reaction a is 74% and reaction b is 59%.