Question

Select the correct answer from each drop-down menu. a reaction proceeds with 2.72 moles of magnesium chlorate and 3.14 moles of sodium hydroxide. this is the equation of the reaction: mg(clo3)2 2naoh → mg(oh)2 2naclo3. determine the theoretical amount of each product that the reaction will produce. the reaction will produce of magnesium hydroxide and of sodium chlorate.

Answer 1

334.22 grams of sodium chlorate and 91.56 grams of magnesium hydroxide would be produced respectively.

Stoichiometric calculation

From the equation of the reaction:

$Mg(ClO_3)_2 + 2NaOH --- > 2NaClO_3 + Mg(OH)_2$

The mole ratio of magnesium chlorate to sodium hydroxide is 1:2.

With 2.72 moles of magnesium chlorate and 3.14 moles of sodium hydroxide provided respectively, the sodium hydroxide is limiting in availability.

Mole ratio of NaOH and $NaClO_3$ = 1:1

equivalent mole of  $NaClO_3$ = 3.14 moles

Mass of 3.14 moles  $NaClO_3$  = 3.14 x 106.44 = 334.22 grams

Mole ratio of NaOH and $Mg(OH)_2$ = 2:1

Equivalent mole of  $Mg(OH)_2$ = 3.14/2 = 1.57 moles

Mass of 1.57 moles  $Mg(OH)_2$ = 1.57 x 58.32 = 91.56 grams

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