How much magnesium sulphide can be obtained from 2 g of magnesium and 2 g of sulphur by the reaction:Mg+S→MgS?Which is the limiting reactant?Calculate the amount of one of the reactant whh remains unreacted.

Divide  the quantities of both the reactants with their respective gram atomic weights.

Then We have 2/24 = 1/12 = 0.08 mols of Magnesium  and 2/32 =1/16 = 0.0625 mols of sulphur .

Then sulphur being in lesser quantities will act as the limiting reagent.

Thus only 0.0625 mols of MgS will be produced having mass 3.5 g and remaing 0.0175 mols of  Mg  having mass 0.5g

mg is limiting reactant

Mg + S = MgS

Limiting reagent is sulphur as its molar mass is higher.

32g of sulphur reacts with 24g Mg

1g of suphur reacts with 24/32g of Mg

Hence, 2g of sulphur reacts with 24/32*2 g of Mg = 1.5 g

Hence 0.5 g of Mg is unreacted.

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