The theoretical amount of each product obtained are:
- 1.57 moles of Mg(OH)₂
- 3.14 moles of NaClO₃
<h3>Balanced equation</h3>
Mg(ClO₃)₂ + 2NaOH —> Mg(OH)₂ + 2NaClO₃
<h3>How to determine the limiting reactant</h3>
From the balanced equation above,
1 mole of Mg(ClO₃)₂ reacted with 2 moles of NaOH
Therefore,
2.72 moles of Mg(ClO₃)₂ will react with = 2.72 × 2 = 5.44 moles of NaOH
From the calculation above, we can see that a higher amount (5.44 moles) of NaOH than what was given (3.14 moles) is needed to react completely with 2.72 moles of Mg(ClO₃)₂
Therefore, NaOH is the limiting reactant
<h3>How to determine the theoretical yield of Mg(OH)₂</h3>
From the balanced equation above,
2 moles of NaOH reacted to produce 1 mole of Mg(OH)₂
Therefore,
3.14 moles of NaOH will react to produce = 3.14 / 2 = 1.57 moles of Mg(OH)₂
Thus, the theoretical yield of Mg(OH)₂ is 1.57 moles
<h3>How to determine the theoretical yield of NaClO₃</h3>
From the balanced equation above,
2 moles of NaOH reacted to produce 2 mole of NaClO₃
Therefore,
3.14 moles of NaOH will also react to produce 3.14 moles of NaClO₃
Thus, the theoretical yield of NaClO₃ is 3.14 moles
Learn more about stoichiometry:
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