if you react one mole of sodium hyrdoxide with excess copper(II) sulfate, how many moles of copper(II) hydroxid are produced?
2NaOH + CuSO4 -> Cu(OH)2 + Na2SO4
Half a mole of copper (II) hydroxide are produced per mole of sodium hydroxide.
To determine the number of moles of copper(II) hydroxide produced when one mole of sodium hydroxide reacts with excess copper(II) sulfate, we need to first balance the chemical equation for the reaction.
The balanced chemical equation for the reaction is:
2NaOH + CuSO4 -> Na2SO4 + Cu(OH)2
From the balanced equation, we can see that for every 2 moles of sodium hydroxide, 1 mole of copper(II) hydroxide is produced.
Given that we have one mole of sodium hydroxide, we can calculate the number of moles of copper(II) hydroxide produced using the ratio from the balanced equation:
1 mole NaOH * 1 mole Cu(OH)2 / 2 moles NaOH = 0.5 moles Cu(OH)2
Therefore, when one mole of sodium hydroxide reacts with excess copper(II) sulfate, 0.5 moles of copper(II) hydroxide are produced.