Given that CaO(s)+ H₂O(l)$\rarr$Ca(OH)₂(s), $\Delta$H°_rxn = -64.8 kJ/mol, how many grams of CaO must react in order to liberate 525 kJ of heat?

Question

Quizplus · Accepted Answer

To solve this problem, we can use the following equation: 
q = n × ΔH 
where q is the amount of heat released (in kJ), n is the number of moles of CaO, and ΔH is the enthalpy of the reaction (in kJ/mol).

We are given that ΔH = -64.8 kJ/mol and that we want to liberate 525 kJ of heat.

Substituting these values, we get:

525 kJ = n × (-64.8 kJ/mol)

Solving for n, we get:

n = -525 kJ / (-64.8 kJ/mol) = 8.11 mol

Therefore, we need 8.11 moles of CaO to liberate 525 kJ of heat. Finally, we use the molar mass of CaO to convert moles to grams:

m(CaO) = n × M(CaO) = 8.11 mol × 56.08 g/mol ≈ 455 g

Therefore, the answer is C (455 g).

Given That CaO(s)+ H2O(l) →\rarr→ Ca(OH)2(s) Δ\DeltaΔ H°rxn = -64

Given That CaO(s)+ H₂O(l) $\rarr$ Ca(OH)₂(s) $\Delta$ H°_rxn = -64