
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.
- 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:
Step 1: Calculate the moles of propane as follows:
Mass of methane = 1 g
Molar mass of methane = 16 g/mol
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:
Step 3: Calculate the mass of H2O formed as follows:
Moles of H2O = 0.125
Molecular weight of H2O = 18 g/mol
2) Combustion of ammonia as follows:
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
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:
Step 3: Calculate the mass of NH3 needed as follows:
Moles of NH3 needed = 0.0833
Molecular weight of NH3 = 17 g/mol
Therefore, mass of ammonia needed is 1.42 g.
Chapter 9 Solutions
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