The equation for the reaction is given as;CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)Volume of CuSO₄ as 46.0 mL;Molarity of CuSO₄ as 0.584 M
We are required to calculate the mass of Cu(OH)₂ precipitatedWe are going to use the following steps;Step 1: Calculate the number of moles of CuSO₄ used
Molarity = Number of moles ÷ Volume
To get the number of moles;
Moles = Molarity × volume
= 0.584 M × 0.046 L
= 0.0269 molesStep 2: Calculate the number of moles of Cu(OH)₂ produced From the equation 1 mole of CuSO₄ reacts to give out 1 mole of Cu(OH)₂Therefore; Mole ratio of CuSO₄ to Cu(OH)₂ is 1 : 1.
Thus, Moles of CuSO₄ = Moles of Cu(OH)₂
Hence, moles of Cu(OH)₂ = 0.0269 molesStep 3: Calculate the mass of Cu(OH)₂
To get mass we multiply the number of moles with the molar mass.
Mass = Moles × Molar mass
Molar mass of Cu(OH)₂ is 97.561 g/mol
Mass of Cu(OH)₂ = 0.0269 moles × 97.561 g/mol
= 2.624 g
Thus, the mass of Cu(OH)₂ that will precipitate is 2.624 g
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