Chemistry, asked by brazilserver12211, 8 days ago

How much volume of nitrogen is required y
to form 11.2 L of NH3 at 273k and 1 atm pressure​

Answers

Answered by vshusharma5839
0

Correct option is

A

67.2 L

Given: Mass of elemental boron = 21.6 g

Temperature = 273 K

Pressure = 1 atm

Solution: The balanced chemical equation for the reduction of boron dichloride is

BCl

3

+1.5H

2

→B+3HCl

From the equation, it is clear that 1 mole of boron formation requires 1.5 moles of hydrogen gas.

Let's calculate the moles of boron from the given mass of boron.

No. of moles =

Molar mass

Mass

=

10.8 gmol

−1

21.6g

= 2.00 mol

So, to obtain 2.00 mol of boron, 3.00 mol of hydrogen gas is required.

One mole of hydrogen gas at 273 K and 1 atm occupies a volume of 22.4L.

Therefore, volume of hydrogen gas = 3.00mol×22.4 Lmol

−1

= 67.2L

Answered by kartikjadhav131006
0

Explanation:

Given: Mass of elemental boron = 21.6 g

Temperature = 273 K

Pressure = 1 atm

Solution: The balanced chemical equation for the reduction of boron dichloride is

BCl

3

+1.5H

2

→B+3HCl

From the equation, it is clear that 1 mole of boron formation requires 1.5 moles of hydrogen gas.

Let's calculate the moles of boron from the given mass of boron.

No. of moles =

Molar mass

Mass

=

10.8 gmol

−1

21.6g

= 2.00 mol

So, to obtain 2.00 mol of boron, 3.00 mol of hydrogen gas is required.

One mole of hydrogen gas at 273 K and 1 atm occupies a volume of 22.4L.

Therefore, volume of hydrogen gas = 3.00mol×22.4 Lmol

−1

= 67.2L

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