33 :
How much magnesium sulphide can be obtained from 2.00 g of magnesium and 2.00 g
sulphur by the reaction, Mg + S-> MgS? Which is the limiting agent? Calculate the amo
of one of the reactants which remains unreacted. [Mg = 24, S = 32]
Answers
Answered by
0
Answer:
Do 24×2+32×2.
the limiting agent is S
Answered by
29
We need to find the Limiting Reagent and the reactant which remains unreacted . You should know that ,
- Limiting Reagent is defined as the reactant which is completely consumed during the reaction.
• Now let's find out the no. of moles :-
• The given reacⁿ is :-
- The given equation is balanced since there are equal no. of atoms on both sides.
- The Sulphur is the limiting reagent , as no. of moles of magnesium is more than that of sulphur. S is limiting reagent.
- The molar mass of MgS is 56 g .
By the equation we see that 1 mole of Sulphur gives 1 mole of MgS. Therefore 0.0625 mole of Sulphur will give 0.0625 mole of MgS .
- Therefore the mass of MgS formed = 56 * 0.0625 = 3.57 g .
• Mass of Mg left :-
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