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
3 answers
I calculated the answer is 58.8%, but i'm not sure. Can anyone please help me make sure this answer please.
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 =
Thank you so much DrBob222. Now I understand.
1 answer