Answer to Question #135657 in General Chemistry for Lois An Macalalad

Question

5.85 Using the balanced equation, 2NO + O2  2NO2, determine the limiting reactant under each of the following conditions.
A. 1.0 mol NO and 1.0 mol O2
B. 2.0 mol NO and 0.50 mol O2
C. 10.0 g NO and 10.0 g O2
D. 28.0 g NO and 16.0 g O2

5.86 For each set of reaction conditions in Problem 5.85, calculate how many moles of the product NO2 form. Then, calculate the number of grams of product that forms.

Answer 1

QUESTION #135657

5.85 Using the balanced equation, 2NO + O2 ® 2NO2, determine the limiting reactant under each of the following conditions.

A. 1.0 mol NO and 1.0 mol O2

B. 2.0 mol NO and 0.50 mol O2

C. 10.0 g NO and 10.0 g O2

D. 28.0 g NO and 16.0 g O2

5.86 For each set of reaction conditions in Problem 5.85, calculate how many moles of the product NO2 form. Then, calculate the number of grams of product that forms.

ANSWER

A. 1.0 mol NO and 1.0 mol O₂

Moles of NO present =1.0 mol

Moles of O₂ present =1.0 mol

From equation the mole ratio of NO : O₂ is 2:1

Moles of NO that would be required to react with all O₂  given is (1×2)÷1=2.0 mols the actual moles of NO present is 1.0 mole which makes it the limiting reactant.

Moles of O₂ that would be required to react with all of the given NO is 1×1/2=0.5 mols the actual moles of O2 present is 1mole.

The moles of of the product formed will be

                      NO : NO₂

Mole ratio          2 : 2    

Moles of NO₂ = (1×2) ÷2= 1.0 mol

Grams of product = moles × molar msass (46)

                               = 1×46=46g

B. 2.0 mol NO and 0.50 mol O₂

Moles of NO present =2.0 mol

Moles of O₂ present =0.50 mol

Moles of NO that would be required to react with all O₂ given is (2×0.5)÷1=1.0 mols the actual moles of NO present is 1.0 mole.

moles of O₂ that would be required to react with all of the given NO is 2×1/2=1.0mols the actual moles of O2 present is 0.5 moles which makes it the limiting reactant.

The moles of of the product formed will be

                      O₂ : NO₂

Mole ratio          1 : 2    

Moles of NO2 = (0.5×2) ÷1= 1.0 mol

Grams of product = moles × molar msass (46)

                               = 1×46=46g

C. 10.0 g NO and 10.0 g O₂

Moles of NO =Mass/molar mass =10/30=0.33mols

Moles of O2 =Mass/molar mass =10/32=0.31mols

Moles of NO that would be required to react with all O₂  given is (0.32×2)÷1=0.64 mols the actual moles of NO present is 0.33 moles which makes it the limiting reactant.

moles of O₂ that would be required to react with all of the given NO is 0.33×1/2=0.165 mols the actual moles of O2 present is 0.32mole.

The moles of of the product formed will be

                      NO : NO₂

Mole ratio          2 : 2    

Moles of NO2 = (0.33×2) ÷2= 0.33 mols

Grams of product = moles × molar msass (46)

                               = 0.33×46=15.18g

D. 28.0 g NO and 16.0 g O₂

Moles of NO =Mass/molar mass =28/30=0.93mols

Moles of O₂ =Mass/molar mass =16/32=0.5mols

Moles of NO that would be required to react with all O₂  given is (0.5×2)÷1=1.0 mols the actual moles of NO present is 0.93 moles which makes it the limiting reactant.

Moles of O₂ that would be required to react with all of the given NO is 0.93×1/2=0.465 mols the actual moles of O2 present is 0.5 moles.

The moles of of the product formed will be

                      NO : NO₂

Mole ratio          2 : 2    

Moles of NO2 = (0.93×2) ÷2= 0.93 moles

Grams of product = moles × molar msass (46)

                               = 0.93×46=42.78g