Question

Assume that the complete combustion of one mole of glucose, a monosaccharide, to carbon dioxide and water liberates 2870 kJ2870 kJ of energy (ΔG°′=−2870 kJ/mol(ΔG°′=−2870 kJ/mol ). If the energy generated by the combustion of glucose is entirely converted to the synthesis of a hypothetical compound X, calculate the number of moles of the compound that could theoretically be generated. Use the value ΔG°′compound X=−54.1 kJ/molΔG°′compound X=−54.1 kJ/mol . Round your answer to two significant figures.

Answer 1

Answer:

number of moles of the compound $\approx$ 53 mole

Explanation:

Given that:

The total energy liberated = - 2870 kJ  ( here , the negative sign typical implies the release of energy due to the combustion reaction)

The equation of the reaction can be represented as:

$\mathbf{C_6H_{12}O_6_{(s)} + 6O_{2(g)} \to 6CO_{2(g)}+6H_2O_{(l)}}$

The energy needed to synthesize 1 mole of compound X  = - 54.1 kJ.mol

Thus;

The total energy = numbers of moles of compound × Energy needed to synthesize  1 mole of compound X

Making the numbers of moles of the compound the subject; we have;

numbers of moles of compound = $numbers \ of \ moles \ of \ compound = \dfrac{total \ energy }{Energy \ needed \ to \ synthesize \ 1 \ mole \ of \ compound \ X}$$numbers \ of \ moles \ of \ compound = \dfrac{-2870 \ kJ }{-54.1 \ kJ/mol}$

number of moles of the compound = 53.04990  mole

number of moles of the compound $\approx$ 53 mole to two significant figure