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

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.