10.0 mol of ammonia gas is injected into a 4.0 L container. At equilibrium 1.2 mol of nitrogen gas is found in the container. The number of moles of ammonia gas left in the container must be which of the following?
N2(g) + 3H2(g) <=====> 2NH3(g) + 94 kJ
a. 8.8 d. 6.4
b. 2.4 e. 9.2
c. 7.6
Well, this seems like a tricky chemistry question. Let me clown around with some numbers and help you out!
According to the balanced equation, 1 mole of nitrogen gas (N2) reacts with 3 moles of hydrogen gas (H2) to produce 2 moles of ammonia gas (NH3).
So, if 1.2 moles of nitrogen gas (N2) are found in the container, that means 1.2/1 = 1.2 moles of ammonia gas (NH3) must have been produced.
Given that 10.0 moles of ammonia gas (NH3) were injected initially, the moles of ammonia gas left in the container would be 10.0 - 1.2 = 8.8 moles.
Therefore, the correct answer is option a. 8.8
Hope that clownish explanation brings a smile to your face!
To find the number of moles of ammonia gas left in the container, we can use the stoichiometry of the balanced equation.
The balanced equation is: N2(g) + 3H2(g) <=====> 2NH3(g)
According to the balanced equation, for every 1 mole of nitrogen gas (N2), 2 moles of ammonia gas (NH3) are formed.
Given that 1.2 mol of nitrogen gas is found in the container at equilibrium, this means that 2 x 1.2 = 2.4 moles of ammonia gas have been formed.
Since 10.0 mol of ammonia gas was initially injected into the container, the number of moles of ammonia gas left in the container is 10.0 - 2.4 = 7.6 mol.
Therefore, the correct answer is c. 7.6.
To solve this question, we need to use the balanced equation and the stoichiometry of the reaction.
From the balanced equation, we can see that 1 mole of nitrogen gas (N2) reacts with 3 moles of hydrogen gas (H2) to produce 2 moles of ammonia gas (NH3).
Given that 10.0 mol of ammonia gas is initially injected into the container, we can set up a proportion to find the moles of nitrogen gas that reacts:
(1.2 mol N2) / (2 mol NH3) = (x mol N2) / (10.0 mol NH3)
Cross multiplying, we get:
1.2 mol N2 = (x mol N2 * 10.0 mol NH3) / 2 mol NH3
Simplifying, we get:
1.2 mol N2 = 5x mol N2 / 2
Solving for x, we get:
x = (1.2 mol N2 * 2) / 5 = 0.48 mol N2
Therefore, the number of moles of ammonia gas left in the container is:
10.0 mol NH3 - 0.48 mol NH3 = 9.52 mol NH3
Rounding to one decimal place, the correct answer is:
e. 9.2
asasas
initial moles:
NH3 = 10
N2 = 0
H2 = 0
equilibrium:
N2 = 1.2 moles from problem.
Therefore, H2 must be 3 x 1.2 = 3.6
NH3 must be 10-(2*1.2) = ??