6.0mol of ammonia gas is injected into a 3.0 mol L container. at equilibrium 1.5 mol of hydrogen gas is found in the container. the number of moles of ammonia gas left in the container must be which of the following?
I's 0.50, not 0.05. The short answer to your question is that I didn't. Don't confuse the mols given in the problem with the value of x in the solution. 1.5 mols is given in the problem as mols H2 at equilibrium; therefore, if it had asked for (H2) we would have written 1.5mol H2/3.0 L = 0.5M. But we let x stand for mols of reactants decomposed or mols products formed (from right to left); so if the problem tells us mols H2 at equilibrium is 1.5 and the equilibrium ICE chart tells us that 1.5 mols is 3x, then we know x = 1.5/3 = 0.5 mol at equilibrum.(And note that 0.5 mol x 3 = 1.5 so we still have that 1.5 mols H2 at equilibrium.) N2 = 0.5 mol at equilibrium and 2x or 2* 0.5 = 1.0 mol NH3 has decomposed. If we had 6.0 to start we must have 5.0 left.
Another way to look at it, and it is just as correct is to say, ok, we have 1.5 mols H2 at equilibrium; therefore, what is NH3? 1.5 mols H2 x (2 mols NH3)/3 mol H2) = 1.00 mol. That means that to form 1.5 mol H2 we must have used 1.00 mol NH3. Than to return to the problem, if we had 6.0 t0 start we must have 5.0 at the end. Some will say the second way is shorter (and perhaps easier to understand) but I like for students to get into the habit of making an ICE chart for almost everything.
What's a 3.0 mol container? There is no choices.
OOPSY I MEANT 3.0L
YES IT IS :)
THANK YOU VERY MUCH !!
why did u chose 05 mol instead of the given which is 1.5 mol
To find the number of moles of ammonia gas left in the container at equilibrium, we need to start by understanding the balanced chemical equation for the reaction involving ammonia and hydrogen gas.
The balanced chemical equation for the reaction between ammonia (NH3) and hydrogen (H2) to form the product, nitrogen (N2), is:
2 NH3(g) ↔ N2(g) + 3 H2(g)
From the equation, we can see that two moles of ammonia (NH3) react to form one mole of nitrogen (N2) and three moles of hydrogen (H2). This means that the stoichiometric ratio between ammonia and hydrogen is 2:3.
Initially, 6.0 moles of ammonia gas (NH3) are injected into the container. At equilibrium, we have 1.5 moles of hydrogen gas (H2) in the container.
Since the stoichiometric ratio between ammonia and hydrogen is 2:3, the amount of hydrogen gas formed should be greater than the amount of ammonia gas consumed. Therefore, we can conclude that less than 2.0 moles of ammonia would have been consumed.
To find out the specific number of moles of ammonia left, we subtract the moles of hydrogen formed (1.5 mol) from the initial moles of ammonia (6.0 mol):
6.0 mol - 1.5 mol = 4.5 mol
Therefore, the number of moles of ammonia gas left in the container must be 4.5 mol.
You didn't give any choices.
..........N2 + 3H2 ==> 2NH3
initial....0.....0.......6.0
change....x.....3x........-2x
equil.....x.....3x.......6.0-2x
3x = 1.5 mol so x = 0.5 mol
Then 6.0 - (2*0.5) = ?
Is 5 mols one of your choices?