Question

$56.0 \,L$ of nitrogen gas is mixed with excess of hydrogen gas and it is found that $20\, L$ of ammonia gas is produced The volume of unused nitrogen gas is found to be _______ $L$

Answer 1

Correct Answer: 46

The correct answer is 46
 
To solve this problem, we can use the balanced chemical equation for the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia gas (NH₃):
N₂ + 3H₂ → 2NH₃
According to the stoichiometry of the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
Given: Volume of nitrogen gas (N₂) = 56.0 L Volume of ammonia gas (NH₃) produced = 20 L
Using the ideal gas law, we can relate the volume of a gas to the number of moles of that gas:
n = PV / RT
where: n is the number of moles of the gas P is the pressure of the gas V is the volume of the gas R is the gas constant (0.0821 L·atm/(mol·K)) T is the temperature in Kelvin
Let's calculate the number of moles of nitrogen gas (N₂):
n(N₂) = (P(N₂) * V(N₂)) / (R * T)
Since the pressure, volume, and temperature are not specified, we can assume they are constant throughout the reaction. Therefore, we can write:
n(N₂) = n(NH₃) * (1 mole N₂ / 2 moles NH₃)
n(N₂) = (20 L * 1 mole N₂) / (2 moles NH₃)
n(N₂) = 10 L
Now, let's calculate the volume of unused nitrogen gas (N₂):
V(unused N₂) = V(N₂ initial) - V(N₂ used)
V(unused N₂) = 56.0 L - 10 L
V(unused N₂) = 46.0 L
Therefore, the volume of unused nitrogen gas is 46.0 L.

 

Answer 2

The correct answer is 46
\(N_2(g)+3H_2(g)→2NH_3(g)\)
Since H₂ is in excess and 20 L of ammonia gas is produced.
Hence, 2 moles NH₃ ≡ 1 mole N₂ (v∝n)
20 L NH₃≡10 L N₂
Volume of N₂ left =56–10
=46 L