Question

Calculate the amount of one of the reactant which left unreacted in the given reaction : $Mg+S\to MgS$ when $12.0gofMg$ reacts with $12.0gofS$. (Given : Molar mass of Mg = 24 g and Molar mass of S = 32 g)
(correct answer +1, wrong answer -0.25)

• A. 3.12 g
• B. 10.12 g
• C. 5.12 g
• D. 4.12 g

Answer 1

The correct option is A 3.12 g

We know that,
$Moles=\frac{Mass}{Molarmass}$
$MolesofMg=\frac{12.0}{24}=0.5$
$MolesofS=\frac{12.0}{32}=0.37$
From the equation,
$Mg+S\to MgS$
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 and Mg is present in excess.
From equation, one mole of S gives one mole of MgS, so, 0.37 mole of S will react with 0.37 mole of Mg to form 0.37 mole of MgS.
$\therefore$
Mole of Mg left unreacted =$0.5-0.37=0.13mol$
Mass of Mg left unreacted =$0.13\times 24g=3.12g$