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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.

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The theoretical amount of each product obtained are:

  • 1.57 moles of Mg(OH)₂
  • 3.14 moles of NaClO₃

Balanced equation

Mg(ClO₃)₂ + 2NaOH —> Mg(OH)₂ + 2NaClO₃

How to determine the limiting reactant

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

How to determine the theoretical yield of Mg(OH)₂

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

How to determine the theoretical yield of NaClO₃

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:

brainly.com/question/14735801

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