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
<h3>What is stoichiometry?</h3>
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: brainly.com/question/1770619?
