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Sagot :
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:
[tex]Mg(ClO_3)_2 + 2NaOH --- > 2NaClO_3 + Mg(OH)_2[/tex]
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 [tex]NaClO_3[/tex] = 1:1
equivalent mole of [tex]NaClO_3[/tex] = 3.14 moles
Mass of 3.14 moles [tex]NaClO_3[/tex] = 3.14 x 106.44 = 334.22 grams
Mole ratio of NaOH and [tex]Mg(OH)_2[/tex] = 2:1
Equivalent mole of [tex]Mg(OH)_2[/tex] = 3.14/2 = 1.57 moles
Mass of 1.57 moles [tex]Mg(OH)_2[/tex] = 1.57 x 58.32 = 91.56 grams
More on stoichiometric calculations can be found here: https://brainly.com/question/27287858
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