The number of water molecules formed when 2.5 g of Ca(OH)­2 reacts with excess HCl is closest to which of the following?Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(ℓ)A.1.2 x 10-1B.6.8 x 10-2C.4.0 x 1022D.2.0 x 1023

First, we need to find the molar mass of Ca(OH)2. The molar mass of Ca is approximately 40.08 g/mol, the molar mass of O is approximately 16.00 g/mol, and the molar mass of H is approximately 1.01 g/mol. Therefore, the molar mass of Ca(OH)2 is 40.08 g/mol + 2*(16.00 g/mol + 1.01 g/mol) = 74.10 g/mol.

Next, we need to find the number of moles of Ca(OH)2 in 2.5 g. We can do this by dividing the mass of the substance by its molar mass. So, the number of moles of Ca(OH)2 is 2.5 g / 74.10 g/mol = 0.0337 mol.

According to the balanced chemical equation, 1 mole of Ca(OH)2 produces 2 moles of H2O. Therefore, 0.0337 mol of Ca(OH)2 will produce 0.0337 mol * 2 = 0.0674 mol of H2O.

Finally, we need to convert the number of moles of H2O to the number of molecules. We can do this by multiplying the number of moles by Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol. So, the number of water molecules formed is 0.0674 mol * 6.022 x 10^23 molecules/mol = 4.05 x 10^22 molecules.

Therefore, the number of water molecules formed when 2.5 g of Ca(OH)2 reacts with excess HCl is closest to 4.0 x 10^22, which is answer choice C.