Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide:
CaO(s) + H₂O(l) → Ca(OH) ₂
A 4.50 g sample of CaO is reacted with 4.34 g of H₂O
O.How many grams of water remain after the reaction is complete?

Question

Quizplus · Accepted Answer

First, we need to balance the chemical equation: $CaO(s) + H_2O(l) \rightarrow Ca(OH)_2$. The molar mass of $CaO$ is approximately 56.08 g/mol, and for $H_2O$, it's about 18.015 g/mol. For $Ca(OH)_2$, it's approximately 74.09 g/mol.From the given masses, we calculate the moles of $CaO$ and $H_2O$ to find the limiting reactant. For $CaO$: $4.50 \, \text{g} \div 56.08 \, \text{g/mol} = 0.0803 \, \text{mol}$. For $H_2O$: $4.34 \, \text{g} \div 18.015 \, \text{g/mol} = 0.2409 \, \text{mol}$.Since the reaction is 1:1, $CaO$ is the limiting reactant. It will completely react, and we need to find out how much $H_2O$ is left. $0.2409 \, \text{mol} - 0.0803 \, \text{mol} = 0.1606 \, \text{mol}$ of $H_2O$ remain unreacted.To find the mass of the remaining $H_2O$, multiply the moles by the molar mass: $0.1606 \, \text{mol} \times 18.015 \, \text{g/mol} = 2.89 \, \text{g}$. Therefore, 2.90 g of water remain after the reaction is complete.

Calcium Oxide Reacts with Water in a Combination Reaction to Produce