Question

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

4NH₃ + 5O₂ ---> 4NO + 6H₂O

When 20.0 g NH₃ and 50.0 g O2 are allowed to react, which is the limiting reagent?

A) NO
B) O₂
C) H₂O
D) no reagent is limiting
E) NH₃


What is the coefficient for O₂ when the following combustion reaction of a hydrocarbon is balanced?

___C₇H₁₄ + ___ O₂ ---> ___ CO₂ + __ H₂O

A) none of these
B) 10
C) 11
D) 21
E) 42

Answer 1

4 NH3 + 5 O2 → 4 NO + 6 H2O

Let's assume for the moment that ammonia is the limiting reactant. Let's calculate the mass of oxygen that would be required to react completely with 20.0 g of ammonia. We can use dimensional analysis and equation coefficients to convert 20.0 g of NH3 to moles of NH3, to moles of O2, to grams of O2. We shall need the following equalities to set up conversion factors:

1 mol NH3 = 17.03052 g NH3
4 mol NH3 = 5 mol O2
1 mol O2 = 31.9988 g O2

[(20.0 g NH3)/1][(1 mol NH3)/(17.03052 g NH3)][(5 mol O2)/(4 mol NH3)][(31.9988 g O2)/(1 mol O2)] = 46.972724 g O2

Answer: Since only 46.972724 grams of oxygen is required, and as much as 50.0 grams of oxygen is available, oxygen is in excess, and ammonia is the limiting reactant indeed

E)Ammonia

2C7H14 + 21O2 -> 14CO2 + 14H2O

D)21

Answer 2