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

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 MgRead more on Sarthaks.com - https://www.sarthaks.com/365839/how-much-magnesium-sulphide-can-obtained-from-00-of-magnesium-and-sulphur-the-reaction-mgs

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 :-

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