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chemistry

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?

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nicole

  1. What's a 3.0 mol container? There is no choices.

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    DrBob222

  2. OOPSY I MEANT 3.0L

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    nicole

  4. 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?

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    DrBob222

  5. YES IT IS :)
    THANK YOU VERY MUCH !!

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    nicole

  6. why did u chose 05 mol instead of the given which is 1.5 mol

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    nicole

  8. 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.

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    DrBob222

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