Chapter 9, Problem 27A

Interpretation Introduction

Interpretation:

The mass of ammonia needed to react with excess oxygen to produce the same amount of water as 1 g of methane, needs to be deduced.

Concept Introduction:

• A chemical reaction is represented in terms of a chemical equation with the reactants on the left and the products on the right.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation, i.e., the stoichiometry gives the amount of reactants and products involved in the reaction.
• Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants.

Answer to Problem 27A

Mass of ammonia = 1.42 g

Explanation of Solution

1) Combustion of methane is represented as follows:

  ${\text{CH}}_4{\text{(g) + 2O}}_{\text{2}}\text{(g)}\to {\text{ CO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(g) }$

Step 1: Calculate the moles of propane as follows:

Mass of methane = 1 g

Molar mass of methane = 16 g/mol

  $\begin{array}{l}\text{Moles of methane = }\frac{\text{Mass of methane}}{\text{Molar Mass methane}}\\ =\frac{1\text{ g}}{16\text{ g/mol}}=\text{0}.0625\text{ moles}\end{array}$

Step 2: Calculate the moles of H₂O formed as follows:

Based on the reaction stoichiometry:

1 mole of methane produces 2 moles of water

Therefore, 0.0625 moles of methane would yield:

  $\frac{0.0625\text{moles methane}\times 2{\text{ moles H}}_2\text{O}}{\text{1 mole methane}}=0.125{\text{ moles H}}_{\text{2}}\text{O}$

Step 3: Calculate the mass of H₂O formed as follows:

Moles of H₂O = 0.125

Molecular weight of H₂O = 18 g/mol

  $\begin{array}{l}{\text{Mass of H}}_{\text{2}}\text{O = Moles }\times \text{ Molecular weight}\\ =0.125\text{ moles }\times \text{ 18 g/mol = }2.25\text{ g}\end{array}$

2) Combustion of ammonia as follows:

  ${\text{4NH}}_3{\text{(g) + 5O}}_{\text{2}}\text{(g) }\to {\text{ 4NO(g) + 6H}}_{\text{2}}\text{O(g)}$

Step 1: Calculate the moles of water corresponding to a mass of 2.25 g as follows:

Mass of water= 2.25 g

Molar mass of water = 18 g/mol

  $\begin{array}{l}\text{Moles of water = }\frac{\text{Mass of water }}{\text{Molar Mass water }}\\ =\frac{2.25\text{ g}}{18\text{ g/mol}}=\text{0}.125\text{ moles}\end{array}$

Step 2: Calculate the moles of NH₃ needed as follows:

Based on the reaction stoichiometry:

4 moles of ammonia reacts to form 6 moles of water

Therefore, moles of ammonia needed to form 0.125 moles of water would be:

  $\frac{4\text{moles ammonia}\times 0.125{\text{ moles H}}_2\text{O}}{{\text{6 moles H}}_2\text{O}}=0.0833{\text{ moles NH}}_3$

Step 3: Calculate the mass of NH₃ needed as follows:

Moles of NH₃ needed = 0.0833

Molecular weight of NH₃ = 17 g/mol

  $\begin{array}{l}{\text{Mass of NH}}_3\text{ = Moles }\times \text{ Molecular weight}\\ =0.0833\text{ moles }\times \text{ 17 g/mol = }1.42\text{ g}\end{array}$

Conclusion

Therefore, mass of ammonia needed is 1.42 g.