[johnsy123]

I don't know how to complete this question. I assume that you use ratios, but i don't where to start. Do you find out which reactant is in excess?

1.5 mol of ammonia was mixed with 0.60 mol of oxygen gas in a sealed vessel and allowed to react completely according to the chemical equation:
4NH3+502 ----> 4NO+6H20
After the reaction was complete, what would the total amount of gas(in mol) in the vessel be?

-cheers

[Nobby]

Change the equation in this way that you have 0.6 mol O₂ in use.

How does the equation looks like.

 x NH₃ + 0,6 O₂ => x NO and y H₂O

please calculate x and y.

[Borek]

How does the equation looks like.

x NH₃ + 0,6 O₂ => x NO and y H₂O

There is no need to do tricks to the coefficients. They should be used to calculate rations, but not by modifying correctly balanced reaction equation.

Johnsy: this is a limited reagent question. Do you know how to read reaction equation? How many moles of oxygen needed for all teh ammonia to react? Do you have that much? If not - what does it mean?

[Nobby]

But this the way to get the ratios. What is wrong with it. Has to be find out what is minimum and maximum of the edukts.  If change to 0.6 O₂ he can calculate that he need 0.48 mol NH₃ and also 0,48 mol NO will  be created. 0,72 mol water will be obtained.
From the given 1.5 mol ammonia he can calculate how much is left. And the summary gives the result.

If you change in the way

1.5 NH₃ + 1.875 O₂ => 1.5 NO + 2.25 H₂O

It is easily to see that it is not enough O₂. We have only 0.6 mol.

[Borek]

But this the way to get the ratios.

Correctly balanced reaction equation has - by definition - lowest integer coefficients, and they define the ratios. Changing coefficients will not change ratios, but it will make the reaction equation non standard. The same coefficients are used in many other places, when dealing with equilibrium and reaction thermodynamics, data tables are always constructed assuming correctly balanced reaction equation, so it doesn't make sense to modify them.

Don't mistake coefficient ratio with amount ratio, even if they have the same numerical value.

[Nobby]

Ok, we talk again not same language.

Please show then he should calculate it. Because I also don't understand to do it then. But I am sure my calculation is correct.

[johnsy123]

Is the answer 1.2.
this is how i done it.

5/4=0.6/x
x=0.48 for nh3

5/6=0.6/y
y=0.72

total produced=0.72+0.48=1.20

which is an option

Note this question is multiple choice

the options are 1, 1.2, 2.2 and 10.

[Nobby]

Yes this is the way how to calculate. But consider the ammonia again!!! 0.48 is consummed, but you used 1.5 mol, what means something left.

[johnsy123]

Sorry the 5/4=0.6/x

x=0.48 for NO not nh3

am i right?

[Nobby]

Yes 0.48 NO, 0.72 H₂O

But you used 1.5 ammonia but consummed only 0.48 of it. So what does it means in your calculation.

[johnsy123]

1.02 left over

[Borek]

Please show then he should calculate it. Because I also don't understand to do it then. But I am sure my calculation is correct.

Check out this page: http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

You don't touch the reaction equation, you use coefficients to calculate ratio and amounts. This is mainly a notational thing and your calculations are correct when it comes to numerical values, just don't write things like

1.5 NH₃ + 1.875 O₂ => 1.5 NO + 2.25 H₂O

as they are confusing. And technically - if treated as reaction equations - they are incorrect by definition.

[Nobby]

Correct 1.02 mol left. So what chemicals now in your vessel and how much. You close to the solution.

[johnsy123]

There is 1.02mol of nh3 left over and 1 mol of o2 left over.

[Nobby]

Ammonia is correct but oxygen is consummed. You used 0.6 mol, where should the 1 mol comes from?

1.5 - 0.48 =1.02 NH₃

0.6-0.6 = 0 O₂

0 + 0.48 = 0.48 NO

0 + 0.72 = 0.72 H₂O

The sum gives 2.2 mol of all gas