How much calcium hydroxide is produced by the complete reaction of 150.0 g CaO with 50.0 g of water?

The reaction between calcium oxide (CaO) and water (H2O) to produce calcium hydroxide (Ca(OH)2) is as follows:

CaO + H2O → Ca(OH)2

From the balanced chemical equation, we can see that one mole of CaO reacts with one mole of H2O to produce one mole of Ca(OH)2.

First, we need to calculate the number of moles for CaO and H2O.

The molar mass of CaO = 40.08 g/mol (for Ca) + 16.00 g/mol (for O) = 56.08 g/mol The molar mass of H2O = 1.01 g/mol (for H) x 2 + 16.00 g/mol (for O) = 18.02 g/mol

Number of moles of CaO = mass / molar mass = 150.0 g / 56.08 g/mol = 2.67 mol Number of moles of H2O = mass / molar mass = 50.0 g / 18.02 g/mol = 2.77 mol

The reaction is limited by the reactant that is present in the smallest amount. In this case, it is CaO because it has fewer moles.

Therefore, the number of moles of Ca(OH)2 produced is equal to the number of moles of CaO, which is 2.67 mol.

Finally, we calculate the mass of Ca(OH)2 produced.

The molar mass of Ca(OH)2 = 40.08 g/mol (for Ca) + 15.999 g/mol (for O) x 2 + 1.01 g/mol (for H) x 2 = 74.10 g/mol

Mass of Ca(OH)2 = number of moles x molar mass = 2.67 mol x 74.10 g/mol = 197.8 g

So, 197.8 g of calcium hydroxide is produced by the complete reaction of 150.0 g CaO with 50.0 g of water.