The following Chemical reaction produces a precipitate in an aqueous solution.
NaOH(aq)+Mg(Cl)2(aq)<->NaCl(aq)+Mg(OH)2(s)
With an equilibrium constant of K=1.3. If NaOH and Mg(Cl)2 are added in stoiciometric ratio and 1.20 mol of NaOH are present at Equilibrium, how much precipitate is produced (in moles)?
To find out how much precipitate is produced in moles, we need to calculate the amount of Mg(OH)2 at equilibrium using the provided information.
First, let's determine the initial moles of NaOH and Mg(Cl)2. Since they are added in a stoichiometric ratio, the initial moles of NaOH would be the same as the initial moles of Mg(Cl)2.
Given that 1.20 mol of NaOH is present at equilibrium, we can conclude that the initial moles of NaOH and Mg(Cl)2 would also be 1.20 mol each.
Now, we need to determine the change in moles of NaOH and Mg(Cl)2 based on the stoichiometry of the reaction. The coefficients in the balanced equation tell us that for every 1 mole of NaOH reacting, 1 mole of Mg(OH)2 will be produced.
Thus, the change in moles of NaOH and Mg(Cl)2 will be -1.20 mol each, as they react completely to form Mg(OH)2.
Now, let's consider the equilibrium position.
The equilibrium constant (K) expression for the given reaction is: K = [NaCl]/[Mg(OH)2]
Given that K = 1.3, and the initial concentration of NaCl is zero, we can set up the following equation:
1.3 = 0/[Mg(OH)2]
Since the equilibrium concentration of Mg(OH)2 is unknown, we'll denote it as "x".
Therefore, we can rewrite the equation as: 1.3 = 0/x
Simplifying the equation, we find that x = 0.
From this, we can conclude that no Mg(OH)2 precipitate is present at equilibrium.
Hence, the amount of precipitate produced (in moles) is 0.