Question

40.0 mL of 0.200 M Sodium Hydroxide solution is mixed with 25.0 mL of 0.300 M Magnesium Chloride solution. Calculate the theoretical yield of Magnesium hydroxide and the molarity of the excess reactant following the reaction

Answer 1

Balanced equation:
2 NaOH + MgCl₂ ===> 2 NaCl + Mg(OH)₂

Moles of NaOH solution = 40 x 0.2 /1000 = 0.008 Moles

Moles of MgCl2 solution = 25 x 0.3 /1000 = 0.0075 Moles

In the reaction NaOH will be consumed 2 equivalent. Hence it is the limiting reagent

0.008 Moles  of NaOH can produce only 0.004 Moles of Mg(OH)2

Mass of 0.004 Moles of Mg(OH)2 = 0.004 x 58.31 = 0.233 gm

Excess reagent is MgCl2

Moles of excess reagent = 0.0075- 0.004 = 0.0035 Moles

Total volume of the solution = 30+25 = 55 ml

Concentration of MgCl2 = 0.0035 x 1000 /55 = 0.06363 M