With an equilibrium constant of k=1.3 if NaOH and MgCL2 are added in stoichiometric ratio and 1.2 mol of NaOH are present at equilibrium, how much precipitate is produced (in moles)?
MgCl, magnesium and Chlorine
Your addendum makes no sense.
That's k = 1.3 for what. Is is Mg(OH)2? If so 1.3 isn't right. You've made a typo surely.
I'm sorry, I forgot the Chemical Reaction
NaOH(aq)+Mg(Cl)2(aq)<->NaCl(aq)+Mg(OH)2(s)
And the 1.3? What about it?
Equilbrium Constant.
To determine the amount of precipitate produced, we need to first write the balanced chemical equation for the reaction between NaOH and MgCl2:
2NaOH + MgCl2 → Mg(OH)2 + 2NaCl
From the balanced equation, we can see that the stoichiometric ratio needed for complete reaction is 2 moles of NaOH to 1 mole of MgCl2.
Given that 1.2 moles of NaOH are present at equilibrium, we can determine the amount of MgCl2 required for the stoichiometric ratio:
1.2 moles NaOH × (1 mole MgCl2 / 2 moles NaOH) = 0.6 moles MgCl2
Therefore, if 0.6 moles of MgCl2 are added in stoichiometric ratio, it means that all the MgCl2 will react completely to form precipitate (Mg(OH)2) according to the balanced equation.
Hence, the amount of precipitate produced is 0.6 moles.