how much magnessium sulphide can be obtained from 2g of magnessium and 2 g sulphur by te reaction ? which is the limiting reagent ? calculate the amount of one of the reactants which remains unreacted?
Asked by pkgupta_nproofings
| 12th May, 2014,
09:56: PM
Mg + S → MgS
1 mol 1 mol 1mol
Moles of Mg = Wt. of Mg / Molar mass of Mg = 2/24 = 0.08 moles
Moles of S = Wt. of S / Molar mass of S = 2/32 = 0.0625 moles
Calculation of limiting reagent:
1 mol of Mg requires S = 1 mol
Therefore 0.08 mol of Mg requires 0.08 mol S
But moles of S actually present = 0.0625 moles
Therefore S is the limiting reagent
Mass of Mg unreacted:
Number of moles of Mg unreacted = 0.08-0.0625= 0.0175 mol
Mass of Mg unreacted = moles x molar mass = 0.0175 x 24 = 0.42g
Mg + S → MgS
1 mol 1 mol 1mol
Moles of Mg = Wt. of Mg / Molar mass of Mg = 2/24 = 0.08 moles
Moles of S = Wt. of S / Molar mass of S = 2/32 = 0.0625 moles
Calculation of limiting reagent:
1 mol of Mg requires S = 1 mol
Therefore 0.08 mol of Mg requires 0.08 mol S
But moles of S actually present = 0.0625 moles
Therefore S is the limiting reagent
Mass of Mg unreacted:
Number of moles of Mg unreacted = 0.08-0.0625= 0.0175 mol
Mass of Mg unreacted = moles x molar mass = 0.0175 x 24 = 0.42g
Answered by Vaibhav Chavan
| 13th May, 2014,
11:03: AM
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