Theoretical yield =0.15 g
% yield = 15%
reaction
C6H12O6 → 2C2H5OH + 2CO2
from above reaction, we get to know that,
1 mole of glucose → 2 mole ethanol
for 4 mole of glucose → 8 mole of ethanol
\frac{55.2}{8(46)} = yield of ethanol
yield of ethanol = 0.15g
% yield = (experimental/theoretical )*100
% yield of ethanol = 0.15 x 100 =15%
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