Question

If 25.0 g of NH₃ and 43.1g of O₂ react in the following reaction, how many grams of NO will be formed?

Answer 1

Answer:

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 25.0 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(25 g)/(17.03 g/mol)

= 1.468 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 35.8 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(35.8 g)/(32 g/mol)

= 1.119 mol

The balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

4 mol of NH3 reacts with 5 mol of O2 for 1.468 mol of NH3, 1.835 mol of O2 is required

But we have 1.119 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

According to balanced equation

mol of NO formed = (4/5)* moles of O2

= (4/5)*1.119 = 0.895 mol

use: mass of NO = number of mol * molar mass

= 0.895*30.01

= 26.86 g

Answer: 26.9 g