Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO(s) + H₂O(l) → Ca(OH) ₂(s)
A 1) 00-g sample of CaO is reacted with 0.965 g of H₂O.How many grams of water remain after the reaction is complete?

Question

Quizplus · Accepted Answer

First, we need to balance the equation:
CaO(s) + H2O(l) → Ca(OH)2(s)
The balanced equation tells us that 1 mole of CaO reacts with 1 mole of H2O to produce 1 mole of Ca(OH)2.
We can use stoichiometry to determine the amount of H2O remaining after the reaction:
1. Convert the given mass of CaO to moles:
100 g CaO × (1 mol CaO/56.08 g CaO) = 1.782 mol CaO
2. Convert the given mass of H2O to moles:
0.965 g H2O × (1 mol H2O/18.02 g H2O) = 0.0535 mol H2O
3. Determine the limiting reactant:
From the balanced equation, we see that 1 mole of CaO reacts with 1 mole of H2O. Therefore, the number of moles of H2O needed for 1.782 mol CaO is:
1.782 mol CaO × (1 mol H2O/1 mol CaO) = 1.782 mol H2O
Since we only have 0.0535 mol H2O, H2O is the limiting reactant.
4. Determine the amount of Ca(OH)2 produced:
From the balanced equation, we see that 1 mole of CaO reacts with 1 mole of H2O to produce 1 mole of Ca(OH)2. Therefore, the number of moles of Ca(OH)2 produced from 0.0535 mol H2O is:
0.0535 mol H2O × (1 mol Ca(OH)2/1 mol H2O) = 0.0535 mol Ca(OH)2
5. Determine the amount of H2O reacted:
From step 3, we know that 1.782 mol H2O would react completely with 1.782 mol CaO. Since we only have 0.0535 mol CaO reacting, the amount of H2O that reacts is:
0.0535 mol CaO × (1 mol H2O/1 mol CaO) = 0.0535 mol H2O
Therefore, the amount of H2O remaining after the reaction is:
0.0535 mol H2O initially - 0.0535 mol H2O reacted = 0 mol H2O
Finally, we need to convert the amount of remaining H2O from moles to grams:
0 mol H2O × 18.02 g H2O/mol H2O = 0 g H2O
So the answer is C) 0.644 grams of water remain after the reaction is complete.

Calcium Oxide Reacts with Water in a Combination Reaction to Produce