Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2O (l) → Ca (OH) 2 (s) A 5.00-g sample of CaO is reacted with 4.83 g of H2O. How many grams of water remain after the reaction is complete? Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2O (l) → Ca (OH) 2 (s) A 5.00-g sample of CaO is reacted with 4.83 g of H2O. How many grams of water remain after the reaction is complete? 3.22 0.00 0.179 0.00991 1.04

3.22 g of water

Explanation:

The statements in the question already suggests that water is the reactant in excess.

Number of moles of calcium oxide = mass of calcium oxide / molar mass of calcium oxide

Number of moles of calcium oxide = 5.00g / 56.0774 g/mol = 0.089 moles

From the reaction equation;

If 1 mole of calcium oxide reacts with 1 mole of water

Then 0.089 moles of calcium oxide reacts with 0.089 moles of water

Mass of 0.089 moles of water = number of moles * molar mass

Molar mass of water = 18.01528 g/mol

Mass of water reacted = 0.089 moles * 18.01528 g/mol

Mass of water reacted = 1.6g of water.

Therefore, mass of water left over = 4.83 - 1.6 = 3.2 g of water.