In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: C 6 H 12 O 6 (s)→2C 2 H 5 OH(l) + 2CO 2 (g) If 5.97 g of glucose are reacted and 1.44 L of CO 2 gas are collected at 293 K and 0.984 atm, what is the percent yield of the reaction.

89.4%

Explanation:

We'll begin by obtaining the actual yield of CO2. This can be obtained calculating the number of mole of CO2 produced from the reaction as follow:

Volume (V) = 1.44 L

Temperature (T) = 293 K

Pressure (P) = 0.984 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =..?

PV = nRT

0.984 x 1.44 = n x 0.0821 x 293

Divide both side by 0.0821 x 293

n = (0.984 x 1.44) / (0.0821 x 293)

n = 0.059 mole

Therefore, the actual yield of CO2 is 0.059 mole.

Next we shall the theoretical yield of CO2. This can be obtained as follow:

First, we shall determine the number of mole in 5.97 g of glucose, C6H12O6.

Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180 g/mol

Mass of C6H12O6 = 5.97 g

Mole of C6H12O6 =?

Mole = mass /molar mass

Mole of C6H12O6 = 5.97/180

Mole of C6H12O6 = 0.033 mole

Now, we can calculate the theoretical yield of CO2 as follow:

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

From the balanced equation above,

1 mole of C6H12O6 produced 2 moles of CO2.

Therefore, 0.033 mole of C6H12O6 will produce = 0.033 x 2 = 0.066 mole of CO2.

Therefore, the theoretical yield of CO2 is 0.066 mole.

Finally, we shall determine the percentage of CO2 as follow:

Actual yield = 0.059 mole

Theoretical yield = 0.066 mole.

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 0.059/0.066 x 100

Percentage yield = 89.4%

Therefore, the percentage yield of the reaction is 89.4%

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