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)/(24)="0.0833 moles of Mg"`
`"2.00 g of S"=(2.00)/(32)="0.0625 moles of S"`
From the reaction, `Mg+SrarrMgS`
It follows that one mole of Mg reacts with one mole of S. We are givenb 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 molar mass of Mg to form 0.0625 mole of MgS.
Molar mass of MgS = 56 g
`therefore" Mass of MgS formed "=0.0625xx56.0=3.5g" of "MgS`
Mole of Mg left unreacted `=0.0833-0.0625` mole of `Mg=0.0208" mole of "Mg`
Mass of Mg of left unreacted = moles of Mg `xx` molar mass of Mg
`=0.0208xx"24 g of Mg"=0.4992" g of Mg"=-.5gMg`