Question

33 :
How much magnesium sulphide can be obtained from 2.00 g of magnesium and 2.00 g
sulphur by the reaction, Mg + S-> MgS? Which is the limiting agent? Calculate the amo
of one of the reactants which remains unreacted. [Mg = 24, S = 32]​

Answer 1

Answer:

Do 24×2+32×2.

the limiting agent is S

Answer 2

We need to find the Limiting Reagent and the reactant which remains unreacted . You should know that ,

• Limiting Reagent is defined as the reactant which is completely consumed during the reaction.

• Now let's find out the no. of moles :-

$\boxed{\begin{array}{c} \dashrightarrow\sf n_{Mg}= \dfrac{2}{24}= \dfrac{1}{12}=\red{0.083\ mole }\\\\\dashrightarrow\sf n_{S}= \dfrac{2}{32}= \dfrac{1}{16}=\red{0.0625\ mole }\end{array}}$

• The given reacⁿ is :-

$\sf \to Mg+ S \longrightarrow MgS$

• The given equation is balanced since there are equal no. of atoms on both sides.
• The Sulphur is the limiting reagent , as no. of moles of magnesium is more than that of sulphur. S is limiting reagent.
• The molar mass of MgS is 56 g .

By the equation we see that 1 mole of Sulphur gives 1 mole of MgS. Therefore 0.0625 mole of Sulphur will give 0.0625 mole of MgS .

• Therefore the mass of MgS formed = 56 * 0.0625 = 3.57 g .

$\sf\to Mg_{(left\ unreacted )}= 0.0833 - 0.0625 = \red{0.0208 \ mol \ of \ Mg }$

• Mass of Mg left :-

$\sf \to Mass_{(Mg\ left)}= n_{(Mg)}\times Molar \ Mass \\\\\sf\to Mass_{(Mg\ left)}= 0.0208 \times 24g \\\\\sf\to Mass_{(Mg\ left)}= 0.49 g \approx \boxed{\red{\sf 0.5g }} .$