Question

The first step in the Ostwald process for producing nitric acid is 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g). If the reaction of 150. g of ammonia with 150. g of oxygen gas yields 87. g of nitric oxide (NO), what is the percent yield of this reaction?options:
a. 77%
b. 100%
c. 49%
d. 33%
e. 62%

Answer 1

Answer:

a. 77%

Explanation:

To solve the exercise, you need to know the limiting reagent. The limiting reagent is one that is consumed first in the chemical reaction. The other reagent is known as an excess reagent, and it is the one left over in a chemical reaction.

To know if ammonia or oxygen is the limiting reagent, we should relate the weights of both compounds in the reaction. We know that 4NH3 (68 g) react with 5O2 (160 g), so we should find out how much NH3 is required to react with 150 g of O2:

160 g O2 _______ 68 g NH3

150 g O2 _______ x = 150 g * 68 g / 160 g = 63.75 g NH3

As we have 150 g of NH3 but only 63.75 g are required, we know that ammonia is the excess reagent, so oxygen is the reagent that limits the reaction and that we should use to calculate the percentage yield of the reaction.

In the reaction we observe that 5O2 (160 g) produces 4NO (120g), so:

160 g O2 ______ 120 g NO

150 g O2 ______ x = 150 g * 120 g / 160 g = 112.5 g NO

If the percent yield of the reaction were 100%, it would produce 112.5 g of NO, however only 87 g of NO were obtained, so we should find out what the percentage of reaction yield was:

112.5 g NO ______ 100%

87 g NO _______ x = 87g * 100% / 112.5 g = 77%

Thus we observe that the reaction had a yield of 77%.

Answer 2

Answer:

The option correct is a.

Explanation:

      1. Calculation of the molecular weights of the substances present in the reaction

MW NH3= AW N+ (AWH)X3= 14 g/mol+ (1g/mol)X3 =17g/mol

MW O2= AW OX2= (16 g/mol)X2 =32g/mol

MW NO= AW N+ AW O= 14 g/mol+ 16 g/mol=30g/mol

Where:

MW: MOLECULAR WEIGHT

AW: ATOMIC WEIGHT

N:  nitrogen; O: oxygen; H:  hydrogen

     2.  Relationship between the moles present and the moles necessary for the reaction to take place

In order to determine the relationship we must calculate the number of moles with the amounts used of each reactant (ammonia and oxygen)

mol NH3=\frac{1mol}{17g}x150g=8.82 mol

mol O2=\frac{1mol}{32g}x150g=4.69 mol

RNH3=\frac{MOL present NH3}{MOL necessary  NH3}=\frac{8.82mol}{4mol} =2.205

RO2=\frac{MOL present O2}{MOL necessary  O2}=\frac{4.69mol}{5mol} =0.938

The limit reagent is molecular oxygen because the amount available is less than that required according to the stoichiometry of the reaction. On the other hand, the amount of ammonia available is approximately double that necessary for the reaction to take place.

    3. Percent yield teoric

We must calculate the theoretical yield based on the limiting reagent (O2)

$g NO= \frac{MW NO}{1} x \frac{mol Theorycal NO}{mol TheorycalO2} x mol O2 present=\frac{30gNO}{1 mol NO} x\frac{4 mol NO}{5mol O2} x4,69 mol O2=112,56gNO$

   4. Percent yield of this reaction

$PERCENT yield= \frac{g NO experiment}{g NO theorycal}x100 =\frac{87g}{112.56g} x100=77 percent$