Limiting Reagent and Excess Reagent JEE Chemistry

❓ Question

Magnesium reacts with sulphur to form magnesium sulphide.

$$\boxed{\text{Mg} + \text{S} \rightarrow \text{MgS}}$$

Given:

• Atomic mass of Mg = 24
• Atomic mass of S = 32
• 2 g of Mg and 2 g of S are taken.

Find:

1. Limiting reagent (L.R.)
2. Mass of excess reagent left after the reaction.

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✍️ Solution

Balanced reaction:

$$\text{Mg}+\text{S}\rightarrow\text{MgS}$$

From the equation,

$$24\text{ g Mg} \rightarrow 32\text{ g S}$$

For 2 g Mg, sulphur required is

$$\frac{32}{24}\times2=\frac{8}{3}\approx2.67\text{ g}$$

But only 2 g S is available.

Therefore,

$$\boxed{\text{Sulphur (S) is the Limiting Reagent}}$$

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Mass of Mg left unreacted

Moles of Mg initially,

$$=\frac{2}{24}=\frac{1}{12}=0.0833\text{ mol}$$

Moles of S initially,

$$=\frac{2}{32}=\frac{1}{16}=0.0625\text{ mol}$$

Since the reaction ratio is 1 : 1, sulphur consumes

$$0.0625\text{ mol Mg}$$

Mg left,

$$0.0833-0.0625=0.0208\text{ mol}$$

Mass of Mg left,

$$0.0208\times24=0.5\text{ g}$$

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✅ Final Answer

→ Limiting reagent: $\boxed{\text{Sulphur (S)}}$

→ Mass of Mg left unreacted: $\boxed{0.5\ \text{g}}$