Question

Take the reaction: NH3 + O2 NO + H2O. In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. (i) Which reactant is the limiting reagent? (ii) How many grams of NO are formed? (iii) How much of the excess reactant remains after the reaction?

Answer 1

Given:

NH3 + O2 -> NO + H2O

3.25 g of NH3 are allowed to react with 3.50 g of O2

To find:

(i) Which reactant is the limiting reagent

(ii) How many grams of NO are formed

(iii) How much of the excess reactant remains after the reaction

Solution:

i) On balancing the given equation,

     4 NH3 + 5O2 –> 4 NO + 6H2O

     Mass of 4 moles of NH3 = 4x(17) = 68g

     Mass of 5 moles of O2 = 5x(32) = 160g

     3.5/3.25 < 160/68

Hence, the limiting reagent is O2.

ii)   O2        NO

   160g   ->  112g

    3.5g   ->    x

        x = 2.45g

Hence, 2.45 grams of NO are formed.

iii)  NH3      O2

   68g   –> 160 g

     x      –> 3.5 g

   x = 1.49 g of NH3

Remaining excess = 3.25–1.49 = 1.76 g

Hence, the remaining excess is 1.76g.

Answer 2

Answer:

(i) $O_{2}$ reactant is the limiting reagent.

(ii) $2.45g$ many grams of NO are formed.

(iii) $1.43$g of $NH_{3}$  of the excess reactant remains after the reaction.

Explanation:

TO FIND:

(i) Which reactant is the limiting reagent?

(ii) How many grams of NO are formed?

(iii) How much of the excess reactant remains after the reaction?

GIVEN:

         $NH_{3} +O_{2} NO+H_{2} O$

       MASS OF $NH_{3}$ IS 3.25g

      MASS OF $O_{2}$ IS 3.50g

  $NH_{3} =17\\O_{2} =32\\H_{2} O=18$

i) THE  balancing  equation IS

                  $4NH_{3} +5O_{2}$→$4NO+5O_{2}$

    Mass of 4 moles of $NH_{3}$= 4x(17) = 68g

    Mass of 5 moles of$O_{2}$= 5x(32) = 160g

             $\frac{3.5}{3.5} < \frac{160}{68}$

Hence, the limiting reagent is $O_{2}$

ii)   Grams of NO are formed are

       $O_{2}$→$N_{O}$

       160g  → 112g

          3.5g  →   x

         $X=\frac{112*3.5}{160}$

         $X=2.45g$

             

2.45g grams of $N_{O}$ are formed.

iii)  $NH_{3}$      $O_{2}$

  68g   → 160 g

    x    → 3.5 g

  x =$\frac{68*35}{160}$

   $X=1.49 g of NH3$

Remaining excess =$3.25-1.49$

                                    =1.76g

Hence, the remaining excess is $1.76g$

1.76g of the excess reactant remains after the reaction

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