Given that CaO(s) + H₂O(l) $\rarr$ Ca(OH)₂(s), $\Delta$H^$\circ$_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

To solve this problem, we need to use the given thermochemical equation and the given ∆H value to determine the amount of CaO that will produce 525 kJ of heat.

First, we need to calculate the amount of heat released when 1 mole of CaO reacts:
∆H = -64.8 kJ/mol
So, when 1 mole of CaO reacts, 64.8 kJ of heat is released.

To produce 525 kJ of heat, we need to determine how many moles of CaO are needed:
mol CaO = 525 kJ ÷ 64.8 kJ/mol
mol CaO = 8.10

So, we need 8.10 moles of CaO to produce 525 kJ of heat.

Finally, we can convert moles to grams:
m(CaO) = mol(CaO) × M(CaO)
m(CaO) = 8.10 mol × 56.08 g/mol
m(CaO) = 455 g

Therefore, the answer is choice C, 455 g of CaO.

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

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