Problem 18 - General Chemistry: Principles and Modern Applications

Question

A commercial method of manufacturing hydrogen involves the reaction of iron and steam. $$ 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g}) $$ (a) How many grams of \(\mathrm{H}_{2}\) can be produced from \(42.7 \mathrm{g}\) Fe and an excess of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (steam)? (b) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are consumed in the conversion of \(63.5 \mathrm{g}\) Fe to \(\mathrm{Fe}_{3} \mathrm{O}_{4} ?\) (c) If \(14.8 \mathrm{g} \mathrm{H}_{2}\) is produced, how many grams of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) must also be produced?

Step-by-Step Solution

Answer

(a) 2.052 g of H2 can be produced from 42.7 g Fe. (b) 27.31 g of H2O is consumed in the conversion of 63.5 g Fe to Fe3O4. (c) If 14.8 g H2 is produced, then you should also obtain 425.73 g of Fe3O4.

Step 1: Determine Molar Mass and Stoichiometric Ratios

From the periodic table, find the molar mass of the elements involved: Iron (Fe) = 55.845 g/mole, Hydrogen (H) = 1.008 g/mole, Oxygen (O) = 15.999 g/mole. Hence, determine the molar mass of water (H2O) and magnetite (Fe3O4). For H2O, it's \(2(1.008) g/mole + 15.999 g/mole = 18.015 g/mole\). For Fe3O4, it's \(3(55.845) g/mole + 4(15.999) g/mole = 231.533 g/mole\). The stoichiometric ratios are also important: 3 moles of Fe react with 4 moles of H2O to produce 1 mole of Fe3O4 and 4 moles of H2.

Step 2: Calculate Quantity of Hydrogen (H2) Produced

To calculate the grams of H2 from 42.7g of Fe, first find the moles of Fe using \(moles = mass / molar mass\) which gives, \(42.7 g/55.845 g/mole = 0.765 moles of Fe\). Now using the stoichiometric ratio of 3 moles Fe : 4 moles H2 in the balanced equation, 0.765 moles of Fe will produce \(0.765*4/3 = 1.02 moles of H2\). Converting moles of H2 into mass gives \(1.02 moles*2(1.008 g/mole) = 2.052 g of H2\).

Step 3: Calculate Quantity of Water (H2O) Consumed

Given 63.5 g of Fe, first find the moles of Fe with \(moles = mass / molar mass\) which gives, \(63.5 g / 55.845 g/mole = 1.137 moles of Fe\). Using the stoichiometric ratio of 3 moles Fe : 4 moles H2O in the balanced equation, 1.137 moles of Fe require \(1.137*4/3 = 1.516 moles of H2O\). Convert moles of H2O into mass for the final answer: \(1.516 moles * 18.015 g/mole = 27.31 g\).

Step 4: Calculate Quantity of Fe3O4 Produced

Given 14.8 g of H2, first find the moles of H2 with \(moles = mass / molar mass\) which gives \(14.8 g / 2(1.008 g/mole) = 7.35 moles of H2\). Using the stoichiometric ratio of 4 moles H2 : 1 mole Fe3O4, 7.35 moles of H2 will produce \(7.35*1/4 = 1.8385 moles of Fe3O4\). Convert moles of Fe3O4 into mass for the final answer: \(1.8385 moles * 231.533 g/mole = 425.73 g\).

Key Concepts

Understanding Chemical Reactions

Chemical reactions are processes where substances, called reactants, transform into different substances, known as products. In the provided exercise, we look at the reaction of iron (Fe) with steam (H₂O), resulting in the production of magnetite (Fe₃O₄) and hydrogen gas (H₂). This transformation is balanced by a specific ratio of reactants and products, which is crucial for calculations.

Using a balanced chemical equation like the one given allows us to understand the precise amount of reactants needed and the amount of products formed. In this reaction, three moles of iron react with four moles of steam to produce one mole of magnetite and four moles of hydrogen gas. This stoichiometric balance is fundamental in performing calculations involving chemical reactions.

The chemical equation showcases how atoms are conserved during the reaction, meaning the same types and numbers of atoms appear in the products as in the reactants.

Role of Molar Mass in Stoichiometry

Molar mass is the mass of one mole of a given substance, expressed in grams per mole (g/mol). It acts as a bridge between the atomic scale and the macroscopic scale, allowing us to convert between the number of atoms/molecules and the mass of a substance.

This is pivotal in stoichiometry, the calculation aspect of chemistry focusing on the quantities of reactants and products in a chemical reaction. In our exercise, we calculate the molar masses of different substances like iron, water, and the resulting compounds.

• For example, for water (H₂O), we add the molar masses of two hydrogen atoms and one oxygen atom to get 18.015 g/mol.
• For magnetite (Fe₃O₄), we sum three iron atoms and four oxygen atoms, totaling 231.533 g/mol.

Understanding these calculations helps you to convert grams into moles and vice versa, a vital step in any stoichiometric calculation.

Stoichiometric Calculations Simplified

Stoichiometric calculations use the ratios in a balanced chemical equation to determine the amounts of reactants and products. Let's break down how this applies to our exercise:

First, you must convert the mass of a substance into moles using its molar mass. For example, if you have 42.7 g of iron, you use its molar mass (55.845 g/mol) to find the number of moles: \[\text{Moles of Fe} = \frac{42.7 \text{ g}}{55.845 \text{ g/mol}} = 0.765 \text{ moles}\]

Using the stoichiometric ratios from the balanced equation, calculate the moles of other substances involved. For instance, from 0.765 moles of iron, you can calculate the moles of hydrogen gas produced using the ratio from the balanced equation (3 moles Fe : 4 moles H₂): \[\text{Moles of H}_2 = 0.765 \times \frac{4}{3} = 1.02 \text{ moles}\]

Finally, convert back to grams if needed. This clear step-by-step approach ensures you understand how amounts are derived from balanced equations, making stoichiometric problems easier to tackle.