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Will a precipitate form if you mix 75.0 mL of a NaOH solution with pOH of 2.58 if it is mixed with 125.0 mL of a 0.0018 M MgCl2 solution

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A precipiatate of mass 0.013 g is formed when you mix 75.0 mL of a NaOH solution with pOH of 2.58 and  125.0 mL of a 0.0018 M MgCl2 solution

What is stoichiometry?

In stoichiometry, calculations are made based on mass - mole or mole - volume relationships. First we must put down the balanced reaction equation; 2NaOH(aq) + MgCl2(aq) -----> Mg(OH)2(s) + 2NaCl(aq)

Now Ionically;

2OH^-(aq) + Mg^2+(aq)  -----> Mg(OH)2(s)

Concentration of OH^- = Antilog (-2.58) = 2.6 * 10^-3 M

Number of moles of OH^- = 2.6 * 10^-3 M * 75/1000 = 0.000195 moles

Concentration of Mg^2+ = 0.0018 M

Number of moles of Mg^2+ =0.0018 M * 125/1000 = 0.000225 moles

Since;

1 mole of Mg^2+ reacts with 2 moles of  OH^-

x moles of Mg^2+ reacts with 0.000195 moles of  OH^-

x = 0.0000975 moles

Mg^2+  is the limiting reactant.

1 mole of Mg^2+ yields 1 mole of the precipitate

0.000225 moles of Mg^2+ yields 0.000225 moles of precipitate.

Hence, a precipitate is formed.

Learn more about precipitate: https://brainly.com/question/1770619?

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