Question

Always show your work. No credit will be given for any answers (even if they are correct) for which work is not shown. 

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 

4NH3 + 502 ---> 4NO + 6H2O 

What is the theoretical yield of water, in moles, when 40.0 g NH3 and 50.0g, O2 are mixed and allowed to react? 

Answer 1

Molar mass of NH3,

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

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 40.0 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(40 g)/(17.03 g/mol)

= 2.348 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 50.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(50 g)/(32 g/mol)

= 1.562 mol

Balanced chemical equation is:

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

4 mol of NH3 reacts with 5 mol of O2

for 2.348 mol of NH3, 2.935 mol of O2 is required

But we have 1.562 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

According to balanced equation

mol of H2O formed = (6/5)* moles of O2

= (6/5)*1.562

= 1.875 mol

Answer: 1.87 mol

Answer 2

Answer 3