Based in the equation of the reaction reaction:
The equation of the reaction is given below:
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.
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.
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.
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