Part D When solving problems involving stoichiometric coefficients, the first st

Question

Part D

When solving problems involving stoichiometric coefficients, the first step is to make sure you have a balanced chemical equation. Then, you determine the limiting reagent by using the coefficients from the balanced equation. You can keep track of the amounts of all reactant and products before and after a reaction using an ICF table. Completing the ICF table will also allow you to determine the limiting reagent, and the amount of product formed assumes that the reaction runs to completion with 100% yield. Parts A and C explore these steps in more detail. Let us consider another reaction.

Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO and H2O that would form when 3.40 mol  NH3 and 4.71 mol  O2 react.

Express the amounts in moles to two decimal places separated by a comma.

Explanation / Answer

Given that ammonia and oxygen are reacting to give NO and water

the balanced chemical reaction is

4NH3 + 5O2 gives 4NO + 6H2O

The reagent consumed completely in a reaction is known as limiting reagent. Given that 3.4 mol ammonia and 4.71 mol oxygen are used in the reaction.

From the balanced chemical equation, it is clear that 1 ammonia can react with 5/4 moles of oxygen or 1.25 moles of oxygen.

the given ratio of moles of oxygen to ammonia = 4.71/3.4 = 1.38 ie oxygen is excess and ammonia is the limiting reagent.

From the chemical equation,

4 moles of ammonia produces 4 moles of NO and 6 moles of water

1 mole ammonia produces 1 mole NO and 1.5 moles water

3.5 mole ammonia produces 3.5 moles NO and 5.25 moles water

moleular weight of NO is 30.01 g/mole

molecular weight of water is 18 g/mole

mass of substance = molecular weight * number of moles

thus

mass of NO = 3.5 * 30.01 = 105.04g

mass of O2 = 5.25 *18 = 94.50 g

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