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? A)6.92 g B)56.1 g C)455 g D)606 g E)3.40 × 10⁴ g

Question

Quizplus · Accepted Answer

First, we need to use the given ΔH°rxn to determine the amount of heat released per mole of CaO reacted:

-64.8 kJ/mol CaO × (1 mol CaO/ 56.08 g CaO) = -1.16 kJ/g CaO

Therefore, for 525 kJ of heat to be released, we need to react:

525 kJ/(-1.16 kJ/g CaO) = -452 g CaO

Since we can't react a negative mass of CaO, we must take the absolute value:

| -452 g CaO | = 452 g CaO

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