Question

How much magnesium sulphide can be obtained from 2.00 g of magnesium and 2.00 g of sulphur by the reaction Mg + S → MgS? Which is the limiting reagent? Calculate the amount of the reactants which remains unreacted.

Answer 1

First of all each of these masses are converted into moles: 

2.00 g of Mg = 2/24.3 = 0.0824 moles of Mg 

2.00 g of S = 2/32 = 0.0624 moles of S 

From the equation, Mg + S → MgS, it follows that one mole of Mg reacts with one mole of S. We are given more moles of Mg than of S. Therefore, Mg is in excess and some of it will remain unreacted when the reaction is over. S is the limiting reagent and will control the amount of product. From the equation we note that one mole of S gives one mole of MgS, so 0.0624 mole of S will react with 0.0624 mole of Mg to form 0.0624 mole of MgS. 

Molar mass of MgS = 56.4 g

∴ Mass of MgS formed = 0.0624 × 56.4 g = 3.52 g of MgS 

Moles of Mg left unreacted = 0.0824 – 0.0624 moles of Mg = 0.0200 moles of Mg 

Mass of Mg left unreacted = moles of Mg × molar mass of Mg = 0.0200 × 24.3 g of Mg = 0.486 g of Mg