To calculate the mass of calcium sulfide (CaS) formed in the given reaction, we first need to determine the molar mass of Ca(OH)₂ and CaS.

Molar mass of Ca(OH)₂:
Ca: 40.08 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
(40.08 + 2(16.00) + 2(1.01)) g/mol = 74.10 g/mol

Molar mass of CaS:
Ca: 40.08 g/mol
S: 32.06 g/mol
(40.08 + 32.06) g/mol = 72.14 g/mol

From the balanced chemical equation, we see that 1 mole of Ca(OH)₂ reacts to form 1 mole of CaS. Therefore, the molar ratio of Ca(OH)₂ to CaS is 1:1.

Given that 48 g of Ca(OH)₂ is present in the reaction, we can calculate the amount of CaS formed:

48 g Ca(OH)₂ (1 mol CaS / 1 mol Ca(OH)₂) (72.14 g CaS / 1 mol CaS) = 48 g CaS

Therefore, 48 g of calcium sulfide (CaS) is formed in the reaction.