Answer to Question #339960 in General Chemistry for Bai

Question

If the percentage yield is 1.92%, what mass of KC7H5O2 can be expected?

Answer 1

C7H8(l) + 2KMnO4(aq) = KC7H5O2(aq) + 2MnO2(s) + KOH(aq) + H2O(l)

mole ratio for the reaction is 1:1

m (KC7H5O2) = x

m (C7H8) = (1 mol C7H8 / 92,14 g/mol) x (1 mol C7H5O2 / 1 mol C7H8) x (121,11 g/mol / 1mol KC7H5O2) = 1,314 g

%yield = actualyield / theoreticalyield

1,92% = 1,314 g / x

x = 1,314 / 0,0192 = 68,44 g

m (KC7H5O2) = 68,44 g