Question
The theoretical yield of Cl2 that could be prepared by mixing 15.0 g of manganese dioxide (MnO2) with 30.0 g of HCl is 12.2 g. The limiting reagent was MnO2. When the reaction was run only 8.82 g of Cl2 was collected. The percent yield of Cl2 is_?
a)58.8 %
b)72.3 %
c)29.4 %
d)81.3 %
e)unknown as there is not enough data to determine the percent yield
a)58.8 %
b)72.3 %
c)29.4 %
d)81.3 %
e)unknown as there is not enough data to determine the percent yield
Answers
Alex
I calculated the answer is 58.8%, but i'm not sure. Can anyone please help me make sure this answer please.
DrBob222
No, if the problem states it correctly, the theoretical yield is 12.2g. You collected only 8.82 g so
%yield = (8.82/12.2)*100 =
%yield = (8.82/12.2)*100 =
Alex
Thank you so much DrBob222. Now I understand.