Cu(OH)2 + HCl = ? уравнение реакции К 120 г раствора гидроксида меди (II) с массовой долей 15% добавили 87 г раствора соляной кислоты (Cu(OH)2 + HCl = ?) с концентрацией 35%. Рассчитайте массу образовавшейся соли, запишите молекулярное и ионное уравнение реакции. Определите, какое из исходных веществ было в избытке.

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вопрос

Molecular equation: Cu(OH)2 + 2HCl = CuCl2 + 2H2O
Ionic equation: Cu(OH)2 + 2H+ + 2Cl- = Cu2+ + 2Cl- + 2H2O

First, let's calculate the amounts of each substance involved in the reaction:

For Cu(OH)2:
120g * 0.15 = 18g

For HCl:
87g * 0.35 = 30.45g

Now, let's determine the limiting reagent:

Calculate the molar masses of Cu(OH)2 and HCl:
Cu(OH)2 = 63.55 + 16 + 1.008*2 = 97.55 g/mol
HCl = 1.008 + 35.45 = 36.458 g/mol

Calculate the number of moles of each substance:
n(Cu(OH)2) = 18g / 97.55 g/mol ≈ 0.184 moles
n(HCl) = 30.45g / 36.458 g/mol ≈ 0.834 moles

Determine the limiting reagent by comparing the moles of each substance:
Cu(OH)2 : HCl = 0.184 : 0.834 ≈ 1 : 4.5

HCl is in excess, Cu(OH)2 is the limiting reagent.

Now, let's calculate the mass of the formed salt (CuCl2):

Calculate the molar mass of CuCl2:
CuCl2 = 63.55 + 2*35.45 = 134.45 g/mol

Determine the theoretical yield of CuCl2:
n(CuCl2) = n(Cu(OH)2) = 0.184 moles
m(CuCl2) = n(CuCl2) M(CuCl2) = 0.184 134.45 ≈ 24.71 g

Therefore, the mass of the formed CuCl2 is approximately 24.71 grams.

In conclusion, Cu(OH)2 was in limiting reagent and HCl was in excess. The molecular equation for the reaction is Cu(OH)2 + 2HCl = CuCl2 + 2H2O, and the ionic equation is Cu(OH)2 + 2H+ + 2Cl- = Cu2+ + 2Cl- + 2H2O.