A 45.0\text{ g} sample of impure glucose containing 80.0\% pure glucose (C_6H_{12}O_6) by … — JAMB UTME

A $45.0\text{ g}$ sample of impure glucose containing $80.0\%$ pure glucose ( $C_6H_{12}O_6$ ) by mass undergoes complete fermentation in the presence of zymase enzyme at suitable conditions. What is the volume of carbon dioxide gas, in $\text{dm}^3$ , released at standard temperature and pressure (STP)?

(Relative atomic masses: $C = 12.0$ , $H = 1.0$ , $O = 16.0$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ )

Answer: 8.96 dm^3

Answer

The volume of carbon dioxide gas released at STP is

8.96\text{ dm}^3

The complete fermentation of glucose is represented by the equation

C_6H_{12}O_6 \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2

. Taking into account the

80.0\%

purity, the mass of active glucose is

0.800 \times 45.0\text{ g} = 36.0\text{ g}

, which corresponds to

\frac{36.0}{180.0} = 0.200\text{ mol}

. Because

1\text{ mol}

of glucose yields

2\text{ mol}

CO_2

0.400\text{ mol}

CO_2

is produced. At STP,

0.400\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 8.96\text{ dm}^3

Step-by-Step Solution

Determine the mass of pure glucose in the impure sample

36.0 g of pure glucose

Only the active pure glucose undergoes fermentation.

Calculate the molar mass of glucose (

C_6H_{12}O_6

)

180.0 g/mol

Needed to convert mass of reactant into molar amount.

Calculate the number of moles of glucose fermented

0.200 mol of glucose

Moles = Mass / Molar mass.

Determine moles of CO2 evolved using reaction stoichiometry

0.400 mol of CO2

Fermentation of 1 mole of hexose sugar produces 2 moles of ethanol and 2 moles of carbon dioxide.

Calculate the volume of CO2 gas at STP

8.96 dm^3

Volume = Moles × Molar volume at STP.

Key Concept

Fermentation Stoichiometry and Molar Yield of Alkanols

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