What is the mass of slaked lime obtained by the reaction of 0.5 mol of calcium oxide with water?

What is the mass of slaked lime obtained by the reaction of 0.5 mol of calcium oxide with water? what amount of hot water substance is consumed if 1.42 g of phosphorus (V) oxide is “dissolved” in it to obtain orthophosphoric acid?

1. Let’s write down the equation for obtaining slaked lime:

CaO + H2O = Ca (OH) 2;

2. find the mass of calcium hydroxide:

m (Ca (OH) 2) = n (Ca (OH) 2) * M (Ca (OH) 2);

M (Ca (OH) 2) = 40 + 17 * 2 = 74 g / mol;

n (Ca (OH) 2) = n (CaO) = 0.5 mol;

m (Ca (OH) 2) = 0.5 * 74 = 37 g.

3. Let’s compose the equation of the reaction that occurs when phosphorus oxide (5) is dissolved in hot water:

P2O5 + 3H2O = 2H3PO4;

4.Calculate the chemical amount of the oxide:

n (P2O5) = m (P2O5): M (P2O5);

M (P2O5) = 31 * 2 + 5 * 16 = 142 g / mol;

n (P2O5) = 1.42: 142 = 0.01 mol;

5.determine the amount of water:

n (H2O) = n (P2O5): 3 = 0.01: 3 = 0.0033 mol.

Answer: 37 g; 0.0033 mol.