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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.

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.

Grade:12th Pass

4 Answers

Swapnil Saxena
102 Points
9 years ago

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

onkar muthe
14 Points
9 years ago

mg is limiting reactant

Rahul Kumar
131 Points
9 years ago

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.

ankitesh gupta
63 Points
8 years ago

 

 

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