Molecular equation: CuO + 2HCl -> CuCl2 + H2O

Ionic equation: CuO + 2H+ + 2Cl- -> Cu2+ + 2Cl- + H2O

To find the mass of copper (II) chloride formed, we need to first calculate the moles of CuO and HCl present in the reaction.

Moles of CuO = 5g / molar mass of CuO
Molar mass of CuO = 63.546 + 15.999 = 79.546 g/mol
Moles of CuO = 5g / 79.546 g/mol = 0.0628 mol

Moles of HCl = 50g / molar mass of HCl
Molar mass of HCl = 1.007 + 35.453 = 36.46 g/mol
Moles of HCl = 50g / 36.46 g/mol = 1.373 mol

Since 2 moles of HCl are needed to react with 1 mole of CuO, HCl is present in excess in this reaction.

The limiting reactant is CuO, so we need to find the moles of CuCl2 formed:

Moles of CuCl2 = Moles of CuO = 0.0628 mol

Now calculate the mass of CuCl2 formed:

Mass of CuCl2 = Moles of CuCl2 x molar mass of CuCl2
Molar mass of CuCl2 = 63.546 + (2 x 35.453) = 134.452 g/mol
Mass of CuCl2 = 0.0628 mol x 134.452 g/mol = 8.454 g

Therefore, 8.454 grams of copper (II) chloride will be formed when 5 grams of copper (II) oxide reacts with 50 grams of hydrochloric acid. Hydrochloric acid will be present in excess in the reaction.