CuO+H2SO4=? уравнение реакции Какие продукты образуются в результате взаимодействия CuO +H2SO4 = ? Надо записать молекулярное и ионное уравнение реакции, а также решить задачу. Вот условие: какая масса сульфата меди (II) образуется при растворении 7 г оксида меди (II) в 130 г серной кислоты? Какое из веществ будет в избытке?
To find the mass of copper (II) sulfate formed, we first need to calculate the number of moles of CuO and H2SO4 present:
Molar mass of CuO: 63.55 g/mol + 16.00 g/mol = 79.55 g/mol Number of moles of CuO = 7 g / 79.55 g/mol = 0.088 moles
Molar mass of H2SO4: 1.01 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol Number of moles of H2SO4 = 130 g / 98.09 g/mol = 1.325 moles
From the balanced equation, we can see that 1 mole of CuO reacts with 1 mole of H2SO4 to form 1 mole of CuSO4. Therefore, the limiting reactant is CuO because it requires less moles to react.
Molar mass of CuSO4: 63.55 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 159.61 g/mol Mass of copper sulfate formed = 0.088 moles * 159.61 g/mol = 14.05 g
Therefore, 14.05 g of copper (II) sulfate is formed when 7 g of copper (II) oxide is dissolved in 130 g of sulfuric acid. The excess reactant is sulfuric acid.
Molecular equation: CuO + H2SO4 -> CuSO4 + H2O
Ionic equation: CuO + 2H+ + SO4^2- -> Cu^2+ + SO4^2- + H2O
To find the mass of copper (II) sulfate formed, we first need to calculate the number of moles of CuO and H2SO4 present:
Molar mass of CuO: 63.55 g/mol + 16.00 g/mol = 79.55 g/mol
Number of moles of CuO = 7 g / 79.55 g/mol = 0.088 moles
Molar mass of H2SO4: 1.01 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol
Number of moles of H2SO4 = 130 g / 98.09 g/mol = 1.325 moles
From the balanced equation, we can see that 1 mole of CuO reacts with 1 mole of H2SO4 to form 1 mole of CuSO4. Therefore, the limiting reactant is CuO because it requires less moles to react.
Molar mass of CuSO4: 63.55 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 159.61 g/mol
Mass of copper sulfate formed = 0.088 moles * 159.61 g/mol = 14.05 g
Therefore, 14.05 g of copper (II) sulfate is formed when 7 g of copper (II) oxide is dissolved in 130 g of sulfuric acid. The excess reactant is sulfuric acid.