React 8 grams of copper (ii) oxide (cuo) with 125 grams of hidrochloric acid (hcl) solution. calculate the mass of the salt and water. the 125 grams is the mass of the hcl solution by means hcl have already dissolved in water. you guys so much ;

1) The mass of the produced salt (CuCl₂) = 13.45 g.

2) The mass of the produced water (H₂O) = 1.8 g.

Explanation:

CuO + 2HCl → CuCl₂ + H₂O,

It is clear that 1 mol of CuO reacts with 2 mol of HCl to produce 1 mol of CuCl₂ and 1 mol of H₂O.

no. of moles of CuO = mass/molar mass = (8.0 g)/(79.545 g/mol) = 0.1 mol.

no. of moles of HCl = mass/molar mass = (125.0 g)/(36.46 g/mol) = 3.43 mol.

∴ 0.1 mol of CuO "limiting reactant" reacts with 0.2 mol of HCl "excess reactant".

1 mol of CuO produces → 1 mol of CuCl₂, from stichiometry.

∴ 0.1 mol of CuO produces → 0.1 mol of CuCl₂.

∴ The mass of the produced salt (CuCl₂) = (no. of moles)(molar mass) = (0.1 mol)(134.45 g/mol) = 13.45 g.

1 mol of CuO produces → 1 mol of H₂O, from stichiometry.

∴ 0.1 mol of CuO produces → 0.1 mol of H₂O.

∴ The mass of the produced water (H₂O) = (no. of moles)(molar mass) = (0.1 mol)(18.0 g/mol) = 1.8 g.