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

Answer:

$Y=63.1\%$

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:

$m_{CO}=57.8gO_2*\frac{1molO_2}{32.0gO_2}*\frac{8molCO}{7molO_2} *\frac{28.01gCO}{1molCO} =57.82gCO$

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

$Y=\frac{36.5g}{57.8g} *100\%\\\\Y=63.1\%$

Regards!