Question

A 100mL sample of 0.1MMgCl2(aq) and a 100mL sample of 0.2MNaOH(aq) were combined, and Mg(OH)2(s) precipitated, as shown by the equation above. If the experiment is repeated using solutions of the same molarity, which of the following changes in volume will double the amount of Mg(OH)2(s) produced?

a. Using the same volume of MgCl2(aq) but twice the volume of NaOH(aq)
b. Using twice the volume of MgCl2(aq) but half the volume of NaOH(aq)
c. Using twice the volume of MgCl2(aq) but the same volume of NaOH(aq)
d. Using twice the volume of MgCl2(aq) and twice the volume of NaOH(aq)

Answer 1

Answer:

Using twice the volume of MgCl2(aq)MgCl2(aq) and twice the volume of NaOH(aq)NaOH(aq)

Explanation:

Answer 2

The amount of Mg(OH)2 is doubled when the volume of the reactants is doubled because they still react in the ratio of 2:1.

The reaction between MgCl2 and NaOH occurs as follows;

MgCl2(aq) + 2NaOH(aq) -----> Mg(OH)2(s) + 2NaCl(aq)

We can see that the mole ratio of the reactants is 1:2.

Initial number of moles of MgCl2 = 100/1000 L × 0.1 M = 0.01 moles

Initial number of moles of NaOH =  100/1000 L × 0.2 M = 0.02 moles

If i double the volume of each reactant;

Number of moles of MgCl2 = 200/1000 L  × 0.1 M = 0.02 moles

Number of moles of NaOH =  200/1000 L  × 0.2 M = 0.04 M

We can see that the amount of Mg(OH)2 is doubled when the volume of the reactants is doubled because they still react in the ratio of 2:1.

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