Question

The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively. If the percentage yield of Cl2 is 72%, what is its actual yield? (1 point) a 40.56 g b 42.17 g c 43.38 g d 46.81 g

Answer 1

c 43.38 g

Step-by-step explanation:

The reaction between MnO2 and HCl can be represented by the following balanced equation:

MnO2 + HCl ---> Cl2 + MnCl2 + H2O

From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.

For the fact that HCl gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess, while the limiting reagent should be MnO2 .

Thus, the theoretical yield of Cl2 will be 60.25 g.

By definition, the percentage yield is given by

% Yield = (Actual Yield) / (Theoretical Yield),

This can be simplified to

Actual Yield = % Yield * Theoretical Yield

Plugging in the given values we have

Actual Yield = 72% * 60.25 = 43.38 g