Question

4g of l2 were made to react with 4g of Mg as per the reaction:Mg+l2>Mgl2 Find out the limiting reagent,name and amount of un reacted reagent and the amount of the product formed.

Answer 1

Answer:

limiting reagent-I2

Unreqcted -Mg2,

amount of product-46.6g

Answer 2

Limiting reagent → Iodine (I₂)

Excess Reagent → Magnesium (Mg)

Amount of excess reagent left behind → 3.624 g Mg

Amount of product formed → 4.448 g I₂

Given,

Mg + I₂ → MgI₂

Mass of I₂ taken = 4 g

Mass of Mg taken = 4 g

To Find,

Limiting reagent

Excess Reagent

Amount of excess reagent left behind

Amount of product formed

Solution,

From the given reaction,

Mg + I₂ → MgI₂

we can infer from the stoichiometry that

1 mole of Mg reacts with 1 mole of I₂ to produce 1 mole of MgI₂

The molar mass of Mg = 24 g

The molecular mass of I₂ = 254 g

Now we will calculate the moles of both the reactants

Number of moles x Molar Mass = Given Mass

Let the number of moles of Mg be 'a' and I₂ be 'b'

For Magnesium (Mg),

a x 24 = 4

a = $\frac{4}{24}$

a = 0.167 mol

For Iodine (I₂),

b x 254 = 4

b = $\frac{4}{254}$

b = 0.016 mol

It can be clearly seen that moles of I₂ are less than the moles of Mg

Hence, I₂ is the limiting reagent and Mg is the excess reagent.

The reaction depends on the limiting reagent.

Since I₂ is the limiting reagent in this case and we know the stoichiometry of the reaction, it can be said that -

Number of Moles of MgI₂ formed = Number of moles of I₂

Number of Moles of MgI₂ formed = 0.016

Also, molecular mass of MgI₂ = 24 + 254 = 278 g

Mass of MgI₂ formed = 0.016 x 278

Mass of MgI₂ formed = 4.448 g

Some mass of Mg would be left unreacted. It can be calculated as follows -

Moles of Mg left behind = 0.167 - 0.016

Moles of Mg left behind = 0.151 mol

Mass of Mg left behind = 0.151 x 24

Mass of Mg left behind = 3.624 g

Thus, I₂ and Mg are the limiting and excessive reagents respectively. Also, 4.448 g is the amount of product formed and 3.624 g is the amount of Mg left behind.

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