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?

5 0

2 years ago

Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with N

Katen [24]

Answer:

Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane

Explanation:

The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:

The answer is (R)-2-chlorobutane.

The reaction take splace through $S_{N} _2$ mechansim and inversion in configuration happens.

7 0

3 years ago