Question

Consider the reaction: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O and ΔH = -2800 kJ. How many kJ of energy are released when 18 grams of glucose are combusted?

Answer 1

-616 kilo joules of energy is released when 40 grams glucose are combusted.

Explanation:

Balance equation of the reaction:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O and ΔH = -2800 kJ

1 mole glucose undergoes combustion to release 2800 kJ of energy

atomic mass of 1 mole of glucose = 180.15 gram/mole

40 grams of glucose will have

number of moles = $\frac{mass}{atomic mass of one mole}$

number of moles = $\frac{40}{180.15}$

                            = 0.22 moles of glucose in 40 gram.

1 mole of glucose when undergoes combustion yields -2800 kilo joules of energy

0.22 moles of glucose when undergoes combustion yields

$\frac{2800}{1}$ = $\frac{x}{0.22}$

x = -616 kilo joules of energy is released when 40 grams glucose undergoes combustion. (minus sign indicates release of energy)