Answer:
1068.8 g of N₂ are required.
We can also say, 1.07 kg of N₂
Explanation:
Let's apply the Ideal Gas Law equation to solve the problem:
Formula is P . V = n . R . T
At STP we know, that P is 1 atm and T is 273.15 K. So let's replace the formula:
1 atm . 855 L = n . 0.082 L.atm /mol.K . 273.15 K
(1atm . 855L) / (0.082 L.atm /mol.K . 273.15 K) = n
38.17 moles = n
Let's convert the moles to mass, nitrogen is a dyatomic gas → N₂
Molar mass N₂ = 28 g/mol
38.17 mol . 28 g/mol = 1068.8 g of N₂ are required.
AT STP the 1 moles of gas is equal to the 22.4 L of gas. The mass of the nitrogen gas in the 855 L tank at STP is 1068 g.
The mass of the Nitrogen gas can be calculate using the formula,
[tex]\bold {w = n \times m}\\[/tex]
Where,
w -mass = ?
n - number of moles = [tex]\bold {=\dfrac {855}{22.4} = 38.169\ moles }[/tex] at STP.
m - molar mass [tex]\bold {N_2}[/tex] =
Put the values in the formula,
[tex]\bold {w = 38.169 \times 28 = 1068.73\ g}[/tex]
Therefore, the mass of the nitrogen gas in the 855 L tank at STP is 1068 g.
To know more about STP,
https://brainly.com/question/24050436
