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

Expert Answer:

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