Question

66.8 g of Cobalt reacted with oxygen and 70.5 g of $C{{o}_{2}}{{O}_{3}}$ was collected after the reaction (below) was completed. Calculate the percent yield. (Molar mass of $Co = 59\operatorname{g}/mol$)
 \[4Co+3{{O}_{2}}\to 2C{{o}_{2}}{{O}_{3}}\]
A. 75%
B. 80%
C. 85%
D. 90%

Answer 1

Hint: The formula of percent yield is the ratio between actual yield and theoretical yield which is then multiplied by a hundred. We have to calculate the theoretical yield.

Complete step by step answer:
The balanced chemical equation of the reaction is given as-
\[4Co+3{{O}_{2}}\to 2C{{o}_{2}}{{O}_{3}}\]
The amount of Cobalt is given but that of Oxygen is not given. This implies that the reaction occurred with the help of atmospheric oxygen and therefore Cobalt will act as the limiting reagent.
Calculating the limiting reagent is important because the concentration or amount of product will depend on it, which will give us the theoretical yield.
Now, the amount of Cobalt is 66.8 g and its molar mass is 59 g/mol. So the number of mole of Cobalt given is-
\[=\dfrac{Given\text{ Mass}}{Molar\text{ Mass}}=\dfrac{66.8}{59}=1.132\operatorname{moles}\]
From the balanced chemical equation, it can be derived that four moles of Cobalt give two moles of Cobalt oxide. So 1.132 moles of Cobalt will give-
$=\dfrac{2}{4}\times 1.132=0.566\operatorname{moles}$
Therefore 0.566 moles of Cobalt oxide are obtained. The molar mass of Cobalt oxide is 166 g. So the amount of Cobalt oxide obtained in grams is-
\[=0.566\times 166 = 93.956\operatorname{g}\]
So the theoretical yield is 93.956 g. Calculating the percent yield-
\[\text{Percent Yield = }\dfrac{\text{Actual yield}}{\text{Theoretical yield}}\text{ x 100}\]
\[=\text{ }\dfrac{70.5}{93.956}\text{ x 100}\]
\[=\text{ 75}\text{.035 }\text{%}\]
So, the correct answer is “Option A”.

Note: The percent yield is the ratio of two similar things. It is a dimensionless quantity.
The above problem could also have been solved if you converted everything to grams instead of moles. There are many methods to solve questions like this, but make sure to check the answer.