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2NH3 + MgSO4 + 2H20 — Mg(OH)2 + (NH4)2SO4
Using the chemical equation above answer the following questions.

c. If I have 4.6 moles of MgSO, and 5.4 moles of NH, which one is the limiting
reactant? You must show your work for credit.

d. What is the greatest amount of Mg(OH) that can be made with 4.6 moles of
MgSO, and 5.4 moles of NH3? You must show your work for credit

e. How many moles of the excess reactant is left over after the reaction has
been completed? You must show your work for credit

Sagot :

mickymike92

Based in the equation of the reaction reaction:

  • NH3 is the limiting reactant.
  • the greatest amount of Mg(OH)2 produced is 2.7 moles.
  • 1.9 moles of the excess reactant are left over.

Which one of the reactants is the limiting reactant?

The equation of the reaction is given below:

  • 2NH3 + MgSO4 + 2H20 — Mg(OH)2 + (NH4)2SO4

From the equation of the reaction 2 moles of NH3 reacts with 1 mole of MgSO4

5.4 moles of NH3 will react with 5.4/2 moles of MgSO4 = 2.7 moles of MgSO4.

Therefore, MgSO4 is in excess and NH3 is the limiting reactant.

What is the greatest amount of Mg(OH) that can be made with 4.6 moles of MgSO, and 5.4 moles of NH3?

Since NH3 is the limiting reactant:

2 moles of NH3 will produce 1 moles of Mg(OH)2

5.4 moles of NH3 will produce 5.4÷2 moles of Mg(OH)2 = 2.7 moles of Mg(OH)2

Therefore, the greatest amount of Mg(OH)2 produced is 2.7 moles.

How many moles of the excess reactant is left over after the reaction has been completed?

MgSO4 is the excess reactant.

2.7 moles of MgSO4 are needed but there are 4.6 moles present.

Moles left over = 4.6 - 2.7 = 1.9 moles

Therefore, 1.9 moles of the excess reactant are left over.

Learn more about about excess reactant at: https://brainly.com/question/25685654

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