Question

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. Thank you guys so much ;)))​

Answer 1

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

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

Step-by-step explanation:

• The balanced chemical equation between CuO and HCl is:

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.

• Firstly, we need to calculate the no. of moles of (8.0 g CuO) and (125.0 g HCl):

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.

• From the balanced equation: CuO reacts with HCl with (1: 2) molar ratio.

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

• To calculate the mass of the produced salt (CuCl₂):

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.

• To calculate the mass of the produced salt (H₂O):

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.