Question

Consider the reaction for the production of NO2 from NO:
2NO(g)+O2(g)→2NO2(g)

-If 85.5 L of O2(g), measured at 34.0 ∘C and 633 mmHg , is allowed to react with 143 g of NO, find the limiting reagent.

-If 98.2 L of NO2 forms, measured at 34.0 ∘C and 633 mmHg , what is the percent yield?

Answer 1

no of moles of NO = weight of NO / molar mass of NO

= 143 g / 30 g/mol

= 4.77 mol

find the no of moles using PV = nRT formula

P = 633 mmHg convert in to atm = 0.833 atm

V = 85.5 L, T = 273 +34 = 307 K, R = 0.0821 L atm / mol K

plug all the values in PV = nRT

0.833 x 85.5 = n x 0.0821 x 307

n = 71.22 / 25.20 = 2.826 mol

now lets see the balanced equation

2NO(g)+O2(g)→2NO2(g)

from this one mole of O2 required 2 moles of NO2

accordingly 2.826 moles of O2 required 2 x 2.826 = 5.652 mol NO required

but actually we have 4.77 mole of NO

so limiting ageny is NO

again from the balanced equation

2 moles of NO2 will give 2 mole of NO

4.77 mole of NO will give 4.77 mol of NO2

theritically 4.77 mole NO should form

find the no of moles of NO2 using PV = nRT

P = 633 mmHg convert in to atm = 0.833 atm

V =98.2 L, T = 273 +34 = 307 K, R = 0.0821 L atm / mol K

plug all the values in PV = nRT

0.833 x 98.2= n x 0.0821 x 307

n = 81.8 / 25.20 = 3.24 mol

now % of yield = (obtained moles of NO2 / theritical moles of NO2) x 100

= (3.24 / 4.77 ) x 100

= 67.92 %