Stoichem of Chemistry

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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!

DATA:

Mass of ammonia = 2.10g

Mass of oxygen= 3.44g

(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=?

SOLUTION:
DETERMINATION OF LIMITING REACTANT:

To determine the limiting reactant, first find out the moles of both reactants and then divide each by it’s coefficient in the balanced equation. The reactant with least number be the limiting reactant.

Moles of ammonia= 2.10/17

  = 0.1235mol

0.1235/4= 0.0308

Moles of oxygen= 3.44/32

  = 0.1075mol

0.1075/5= 0.0215

So, oxygen is the liminting reactant

GRAMS OF NO

5mol of O2 produces 4 mol of NO

0.1075mol O2 produces (4*0.1075)/5

  = 0.086mol NO

Mass of NO = mol of NO* molecular mass NO

  = 0.086* 30

  = 2.58g

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


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

I thought it was 1.462 used and 0.638 remaining?

Jul 24th, 2015

Hello?

Jul 24th, 2015

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