Question

In the following unbalanced equation, how many moles of nitrogen monoxide are needed to react with excess ammonia to produce 32.6 grams of water?
NH3 + NO --> N2 + H2O

Answer 1

you need to balance it first

Answer 2

6 NO + 4NH3 --> 5N2 + 6H2O

Answer 3

so now that it is balanced, what do you do next?

Answer 4

Well, it is my mistake. But, no, I don't believe it changes anything.

Answer 5

4 NH3 + 6 NO --> 5 N2 + 6 H2O

Answer 6

The molar mass of water is 18 grams. So, I divided the given mass of water produced by the molar mass of water

Answer 7

32.6 grams produced / 18 grams H2O = 1.81
Is this right?

Answer 8

why?

Answer 9

Why what?

Answer 10

why did you do the steps that you did?

Answer 11

I prefer seeing dimensional analysis involved in the calculation.

Answer 12

I could not figure out another way.

Answer 13

I realize from your response that I clearly am wrong. That is why I was looking for help at all.

Answer 14

Show me all of your accounting first. You have to consider all the masses first hand and then calculate.

Answer 15

All of the masses of each reactant and product in the equation?

Answer 16

If you know how i should set up this equation, I would be very grateful for your help.

Answer 17

well let us make it a little easier by looking at the ratio
for every 4 NH3, you produce 6 H2O

what is the amu of 6 moles H2O?

Answer 18

108 g/mol

Answer 19

and what is the amu of 4 moles of NH3?

Answer 20

68.124 g/mol

Answer 21

you are being asked for how many grams of water to produce?

Answer 22

What about NO?

Answer 23

see now you why you have to account everything in the balanced equation?

Answer 24

The amount of water produced is 32.6 grams

Answer 25

so set up an algebraic ratio based on NH3 and H2O

Answer 26

So, how should I implement the masses to solve?

Answer 27

just do it for now

Answer 28

Do it how?

Answer 29

108/64.124 = x/32.6

Answer 30

But the question asks how many moles of NO are needed

Answer 31

okay now you understand why we have to calculate all the masses in the reactants and products

Answer 32

so let us set up all the masses of our balanced reactants and product

Answer 33

I appreciate your time and your helping me with this.

Answer 34

it's how my professor taught me. He would go along with how I thought to solved it and make me realize why it is wrong

Answer 35

Okay, so I am trying to find the moles of NO needed to yield 32.6 grams of H2O.

Answer 36

So why would I need to divide the mass of H2O by the mass of NH3?

Answer 37

you don't

Answer 38

Oh, I thought that you set up your equation by dividing the mass of H2O (108) by the mass of NH3 (68.124)?

Answer 39

because you didn't want to account all the masses in the reactants
you have to realize that the water produced is coming from ammonia and nitrogen monoxide

Answer 40

So how would I set up the equation to find the number of moles of NO needed?

Answer 41

The mass of 4 moles of NH3 is 68.124 g/mol; 6 moles of NO is 180.0366 g/mol.
6 moles of H2O = 108 g/mol. 10 moles of N is 140.067 g/mol.

Answer 42

Are you still there?

Answer 43

4 NH3 + 6 NO --> 5 N2 + 6 H2O
--------------------------------

How many moles of H2O is 32.6 g of H2O?

Answer 44

1.81

Answer 45

That is what I had initially thought to be the answer to the question. But stoichiometry is too complicated for it to be that simple. Right?

Answer 46

What should I do next?

Answer 47

read up a little bit to get a grasp of the concept of stoichiometry on limiting reagent and excess
http://finedrafts.com/files/chemistry/General/Zumdahl%20etal/8th%20ed/CH03%20Stoichiomestry.pdf

Answer 48

I know how to solve stoichimetry problems to find limiting reagent, theoretical and percent yields, but only when the question gives me a given mass of the reactants. I just want to know how to solve this problem.

Answer 49

Please help me

Answer 50

great. that's good news
if you can calculate the masses of reactants and products in chemical reaction, then you just need a little dimensional analysis in order to find the mass of the reactants in order to produce a desired mass in the chemical reaction

Answer 51

Would the law of conservation of mass in chemical reactions help me solve this using molar masses of the products and reactants?

Answer 52

I have no idea how I will solve this.

Answer 53

how you would normally solve this when finding for the limiting reagent?

Answer 54

I would have been given the mass in grams of NO and NH3 so that I could determine which produced fewer moles of product (in this case, H2O).

Answer 55

you look for the limiting reagent and then use it as the basis of your calculation, yes?
well this is the same, except that you have your work cut out for you. the limiting reagent has been identified

Answer 56

Well yes, the question tells me that the limiting reactant is NO, as NH3 is in "excess." So how should I use what I am given?

Answer 57

Am i missing something?

Answer 58

What should I try to solve for?

Answer 59

The answer is definately not 1.81 moles?

Answer 60

Here is how I solved it:
1.81 mol H2O * 6 mol NO / 6 mol H2O = 1.81 mol NO

Answer 61

I welcome your corrections!

Answer 62

good now you're using dimensional analysis
1.81 mol of H2O x (6 mol of NO/6 mol of H2O) = ___ mol of NO

Answer 63

1.81?