Answer:
The % yield is 74.45 %
Explanation:
<u>Step 1:</u> The balanced equation
C6H12O6⟶2CH3CH2OH+2CO2
<u>Step 2</u>: Data given
Molar mass glucose = 180.15 g/mol
Molar mass of ethanol = 46.08 g/mol
Molar mass of carbon dioxide = 44.01 g/mol
Mass of glucose = 61.5 grams
Mass of ethanol = 23.4 grams
<u>Step 3:</u> Calculate moles of glucose
Moles glucose = Mass glucose / Molar mass of glucose
Moles glucose = 61.5 grams / 180.15 g/mol
Moles glucose = 0.341 moles
<u>Step 4:</u> Calculate moles of ethanol
1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2
0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol
<u>Step 5:</u> Calculate mass of ethanol
Mass ethanol = moles ethanol * Molar mass ethanol
Mass ethanol = 0.682 moles * 46.08 g/mol
Mass ethanol = 31.43 grams = theoretical mass
<u>Step 6:</u> Calculate % yield
% yield = actual mass / theoretical mass
% yield = (23.4 grams / 31.43 grams) * 100%
% yield = 74.45 %
The % yield is 74.45 %
