Question

7.0 of Nitrogen gas is combined with 3.0g of oxygen gas. 4.5g of nitrous oxide is formed during the experiment. The balanced reaction is : N2(g) + O2(g) = 2NO(g).

A. What is the limiting reagent? justify your answer.

B. How many grams of NO is it possible to form (theoretical yield)?

C. what is the percent yield?

Answer 1

Number of moles of N2 = 7.0 g / 28.0 g/mol = 0.25 mole

Number of moles of O2 = 3.0 g / 32.0 g/mol = 0.0938 mole

From the balanced equation we can say that

1 mole of N2 requires 1 mole of O2 so

0.25 mole of N2 will require

= 0.25 mole of N2 *(1 mole of O2 / 1 mole of N2)

= 0.25 mole of O2

But we have 0.0938 mole of O2 which is in short so O2 is limiting reactant

The limiting reactant is: O₂

From the balanced equation we can say that

1 mole of O2 produces 2 mole of NO so

0.25 mole of O2 will produce

= 0.25 mole of O2 *(2 mole of NO / 1 mole of O2)

= 0.50 mole of NO

mass of 1 mole of NO = 30.01 g so

the mass of 0.50 mole of NO = 15.0 g

Therefore, theoretical yield of NO = 15.0 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (4.5 / 15.0)*100

percent yield = 30 %