Question

For the following reaction: 4C2H3Cl + 702 → 8CO + 6H20 + 2Cl2 If the reaction of 57.8 grams of O2 produces 36.5 grams of CO, what is the percent yield?​

Answer 1

`Y=63.1\%`

Step-by-step explanation:

Hello there!

In this case, according to the given chemical reaction:

`4C_2H_3Cl + 7O_2 \rightarrow 8CO + 6H_2O + 2Cl_2`

It turns out possible for us to realize about the 7:8 mole ratio of O2 to CO, and therefore, the theoretical yield of the latter is calculated via stoichiometry:

`m_(CO)=57.8gO_2*(1molO_2)/(32.0gO_2)*(8molCO)/(7molO_2) *(28.01gCO)/(1molCO) =57.82gCO`

Finally, the percent yield is calculated by dividing the actual yield, 36.5 g by the just computed theoretical one:

`Y=(36.5g)/(57.8g) *100\%\\\\Y=63.1\%`

Regards!