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