Question

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

NO:
NH3 + O2 + NO + H2O
balanced:4NH3 +502 UNO+ 6H2O

a.) How many grams of NO form when 1.5g of NH3 reacts with 1.85g of O2? b.) Which
reactant is the limiting reactant and which one is the excess reactant? c.) How much of
the excess reactant remains after the limiting reactant is completely consumed?​

Answer 1

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Step-by-step explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

`\text{moles of NH}_(3) = \text{1.5 g NH}_(3) * \frac{\text{1 mol NH}_(3)}{\text{17.03 g NH}_(3)} = \text{0.0881 mol NH}_(3)\\\\\text{moles of O}_(2) = \text{1.85 g O}_(2) * \frac{\text{1 mol O}_(2)}{\text{32.00 g O}_(2)} = \text{0.057 81 mol O}_(2)`

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

`\text{Moles of NO} = \text{0.0881 mol NH}_(3) * \frac{\text{4 mol NO}}{\text{4 mol NH}_(3)} = \text{0.0881 mol NO}`

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

`\text{Moles of NO} =  \text{0.057 81 mol O}_(2)* \frac{\text{4 mol NO}}{\text{5 mol O}_(2)} = \text{0.046 25 mol NO}`

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

`\text{Mass of NO} = \text{0.046 25 mol NO}* \frac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}`

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

`\text{Moles reacted} = \text{0.057 81 mol O}_(2) * \frac{\text{4 mol NH}_(3)}{\text{5 mol O}_(2)} = \text{0.046 25 mol NH}_(3)`

7. Calculate the mass of NH₃ reacted

`\text{Mass reacted} = \text{0.046 25 mol NH}_(3) * \frac{\text{17.03 g NH}_(3)}{\text{1 mol NH}_(3)} = \text{0.7876 g NH}_(3)`

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃