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

Explanation

We shall convert the masses into moles
2.00 g of `Mg = (2.00)/(224)=0.0833` moles of Mg
2.00 g of `S=(2.00)/(32)=0.0625` moles of S
From the equation,
`Mg+S rarr MgS`
It follows that one mole of Mg reacts with one mole of S. We are given more mole of Mg than S, therefore S is the limiting reagent. From equation, one mole of S gives one mole of MgS, so, 0.0625 mole of S will react with 0.0625 mole of Mg to form 0.0625 mole og MgS.
Molar mass of MgS = 56 g
`therefore` Mass of MgS formed `=0.0625xx56.0 g = 3.5 g` of MgS
Mole of Mg left unreacted `= 0.0833 - 0.0625` mole of Mg = 0.208 mole of Mg
Mass of left unreacted = mole of Mg `xx` molar mass of Mg
`=0.0208xx24g` of `Mg=0.4992 g` of `Mg ~~ 0.5 g Mg`