How many grams of Cu(OH)2 will precipitate when excess NaOH solution is added to 51.0 mL of 0.797 M CuSO4

Question

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Answer 1

First, write the correctly balanced equation for the reaction taking place:

2NaOH(aq) + CuSO₄(aq) ==> Na₂SO₄(aq) + Cu(OH)₂(s) ... balanced double replacement reaction

Next, we find the moles of CuSO₄ that were used:

51.0 ml x 1 L/1000 ml x 0.797 mol/L = 0.04065 moles CuSO₄

Next, using the balanced equation, we find moles of Cu(OH)₂ formed from 0.04065 moles of CuSO₄:

0.04065 moles CuSO₄ x 1 mol Cu(OH)₂ / mole CuSO₄ = 0.04065 moles Cu(OH)₂ formed

Finally, using the molar mass of Cu(OH)₂, we convert moles to grams:

0.04065 mol Cu(OH)₂ x 97.56 g Cu(OH)₂ /mole = 3.97 g Cu(OH)₂ (3 sig. figs.)