Question

the volume of the Methane evolved by treatment of 16.6 gram of methyl magnesium iodide with water at STP is

Answer 1

Answer:

2.24

Explanation:

Answer 2

The volume of the Methane evolved by treatment of 16.6 grams of methyl magnesium iodide with water at STP is 2.253L.

Given:

Methane evolved by treatment of 16.6 grams of methyl magnesium iodide with water.

To Find:

The volume of the Methane evolved by treatment of 16.6 grams of methyl magnesium iodide with water at STP.

Solution:

To find the volume of the Methane evolved by treatment of 16.6 grams of methyl magnesium iodide with water at STP we will follow the following steps:

As we know,

Reaction:

CH3MgI → CH4 + Mg(OH)I

Mass of methyl magnesium iodide = 16.6g

The molecular mass of methyl magnesium iodide = 165g

Number of moles =

$\frac{given \: mass}{molecular \: mass} = \frac{16.6}{165}$

1 mole of methyl magnesium iodide gives 1 mole of methane.

So,

$\frac{16.6}{165}$

a mole of methyl magnesium iodide =

$\frac{16.6}{165}$

a mole of methane.

Now,

At STP, 1 mole of any gas is 22.4L

So,

The volume of methane at STP =

$\frac{16.6}{165} \times 22.4 = 2.253l$

Henceforth, the volume of the Methane evolved by treatment of 16.6 grams of methyl magnesium iodide with water at STP is 2.253L.

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