Chemistry Stoichem Problems

Chemistry
Tutor: None Selected Time limit: 1 Day

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO.

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
In a certain experiment, 2.10 g of NH3 reacts with 3.44 g of O2.

(a) Which is the limiting reactant?

(b) How many grams of NO form?

(c) How many grams of the excess reactant remains after the limiting reactant is completely consumed?

Jul 24th, 2015

Thank you for the opportunity to help you with your question!

Oxygen is the limiting reactant.

2.58g of NO form.

1.978g of NH3 remaining.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 24th, 2015

How did you get the last value of remaining nh3


Jul 24th, 2015

MASS OF AMMONIA REMAINING

5mol of O2 produces 4 mol of NH3

0.1075mol O2 produces (4*0.1075)/5

  0.086 mol NH3

Mass of ammonia= 0.086*17

  = 1.462g

Thats what i got :/

Jul 24th, 2015

We first calculate how much of the reactant was used

3.44g of O2*1mol O2/32g of O2*4mol NH3/5mol O2*17g NH3/1mol NH3= 1.462g used

Then we subtract from the original, which is 2.10

2.10g-1.462g=0.638g used.

I am sorry I was in a hurry and subtracted from the amount of oxygen. Thanks for asking.

In case you need any further help, I am here for you.

Jul 24th, 2015

1.462 USED AND 0.638 REMAINED

Jul 24th, 2015

Thats okay you still did a great job :) So just to be clear the final answer is 1.462g?

Jul 24th, 2015

Oh okay haha thanks again 5 stars :)

Jul 24th, 2015

Thank you so much and you are welcome again. I will be happy to help any time.

Jul 24th, 2015

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