Question

A gummy bear contains 5.24 g of sucrose (342.3 g/mol). When the gummy bear reacts with potassium chlorate (KClO₃) has a standard enthalpy change of −5960 kJ when 1 mol of sucrose reacts with 8 mol of KClO₃ such that the following balanced equation can be generated: C₁₂H₂₂O11(aq)+8KClO₃(aq)→12CO₂(g)+11H₂O(i)+8KCl(aq) If 15.9 g of potassium chlorate are used what is the change in enthalpy per mole of water produced?

a. −1020 kJ/mol
b. −271 kJ/mol
c. −512 kJ/mol
d. −542 kJ/mol
e. 256 kJ/mol
f. 1080 kJ/mol

Answer 1

The change in enthalpy per mole of water produced from the reaction of sucrose with potassium chlorate is -542 kJ/mol, based on the stoichiometric calculations from the given balanced chemical equation.

Step-by-step explanation:

To find the change in enthalpy per mole of water produced, we'll first determine the moles of potassium chlorate used. Then we'll calculate the moles of water produced using stoichiometry of the reaction equation and finally calculate the enthalpy change per mole of water produced.

The molar mass of KClO₃ is approximately 122.55 g/mol. Using this, the moles of KClO₃ that react can be calculated:

15.9 g KClO₃ × (1 mol KClO₃ / 122.55 g KClO₃) = 0.1297 mol KClO₃

Since the stoichiometry of the reaction shows that 1 mol of sucrose reacts with 8 mol of KClO₃ to produce 11 mol of H₂O, we can set up the following ratio:

(0.1297 mol KClO₃ / 8 mol KClO₃) = x mol H₂O / 11 mol H₂O

Solving for x gives us 0.1791 mol H₂O.

The given enthalpy change for the full reaction with 1 mol of sucrose is -5960 kJ. Since 11 mol of water is produced from the complete reaction of 1 mol of sucrose, the enthalpy change per mole of water can be calculated:

-5960 kJ / 11 mol H₂O = -542 kJ/mol H₂O

Therefore, the change in enthalpy per mole of water produced is -542 kJ/mol, which corresponds to option d.