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Industrially, hydrogen gas can be prepared by combining propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) with steam at about \(400^{\circ} \mathrm{C}\). The products are carbon monoxide (CO) and hydrogen gas \(\left(\mathrm{H}_{2}\right) .\) (a) Write a balanced equation for the reaction. (b) How many kilograms of \(\mathrm{H}_{2}\) can be obtained from \(2.84 \times 10^{3} \mathrm{~kg}\) of propane?

Short Answer

Expert verified
The reaction produces 1301.32 kg of hydrogen.

Step by step solution

01

Write the Balanced Chemical Equation

The reaction involves propane (\( \text{C}_3\text{H}_8 \)) reacting with steam (\( \text{H}_2\text{O} \)) to produce carbon monoxide (CO) and hydrogen gas (\( \text{H}_2 \)). Start by writing the unbalanced equation: \[ \text{C}_3\text{H}_8 + \text{H}_2\text{O} \rightarrow \text{CO} + \text{H}_2 \]Balance the carbon atoms:\[ \text{C}_3\text{H}_8 + \text{H}_2\text{O} \rightarrow 3\text{CO} + \text{H}_2 \]Balance the hydrogen atoms:\[ \text{C}_3\text{H}_8 + 4\text{H}_2\text{O} \rightarrow 3\text{CO} + 10\text{H}_2 \]Finally, balance the oxygen atoms:\[ \text{C}_3\text{H}_8 + 4\text{H}_2\text{O} \rightarrow 3\text{CO} + 10\text{H}_2 \]Verify that all atoms are balanced. The balanced equation is:\[ \text{C}_3\text{H}_8 + 4\text{H}_2\text{O} \rightarrow 3\text{CO} + 10\text{H}_2 \]
02

Convert Propane Mass to Moles

To find how many kilograms of \( \text{H}_2 \) can be produced, first convert the mass of propane to moles. The molar mass of \( \text{C}_3\text{H}_8 \) is approximately 44.1 g/mol. Convert the mass from kilograms to grams: \[ 2.84 \times 10^3 \text{ kg} = 2840000 \text{ g} \] Calculate the moles of \( \text{C}_3\text{H}_8 \): \[ \text{Moles of } \text{C}_3\text{H}_8 = \frac{2840000 \text{ g}}{44.1 \text{ g/mol}} \approx 64421 \text{ mol} \]
03

Calculate Moles of Hydrogen Produced

From the balanced equation, \( 1 \text{ mol} \, \text{C}_3\text{H}_8 \) generates \( 10 \text{ mol} \, \text{H}_2 \). Therefore, the moles of hydrogen produced is:\[ \text{Moles of } \text{H}_2 = 64421 \text{ mol} \times 10 = 644210 \text{ mol} \]
04

Convert Moles of Hydrogen to Kilograms

Now, convert moles of \( \text{H}_2 \) to kilograms. The molar mass of \( \text{H}_2 \) is about 2.02 g/mol.Calculate the mass of \( \text{H}_2 \):\[ \text{Mass of } \text{H}_2 = 644210 \text{ mol} \times 2.02 \text{ g/mol} \approx 1301322.2 \text{ g} \]Convert it to kilograms:\[ \text{Mass of } \text{H}_2 = \frac{1301322.2 \text{ g}}{1000 \text{ g/kg}} = 1301.32 \text{ kg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Molar mass is a fundamental concept in chemistry that allows us to convert between the mass of a substance and the amount in moles. It is defined as the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol). To find the molar mass of a compound, sum the atomic masses of all the atoms present in its chemical formula. For example, propane (C\(_3\)H\(_8\)) has a molar mass calculated as follows:

\[ \text{Molar Mass of } \text{C}_3\text{H}_8 = (3 \times 12.01 \, \text{g/mol}) + (8 \times 1.01 \, \text{g/mol}) = 44.1 \, \text{g/mol} \]

This value means that one mole of propane weighs 44.1 grams. Molar mass acts as a bridge between the mass of a substance and its amount in moles, facilitating stoichiometric calculations in chemical reactions.
  • Identify the chemical formula.
  • Use the periodic table for atomic masses.
  • Calculate by adding atomic masses.

This essential concept prepares us for further stoichiometric analysis and quantitative predictions in reactions.
Chemical Reaction Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It relies on a balanced chemical equation to make these predictions. Let's consider the reaction of propane with steam as an example. The balanced equation is:

\[ \text{C}_3\text{H}_8 + 4\text{H}_2\text{O} \rightarrow 3\text{CO} + 10\text{H}_2 \]

This equation tells us that one mole of propane reacts with four moles of water to produce three moles of carbon monoxide and ten moles of hydrogen gas. The coefficients in the balanced equation represent the molar ratios of the reactants and products, which are the key to solving stoichiometric problems.
  • Identify molar ratios from the equation.
  • Use ratios to convert moles of substances.
  • Apply to mass to moles conversions.

The stoichiometric calculations ensure we can predict the amount of product formed from given reactants, aiding in practical applications like industrial processes.
Hydrogen Gas Production
Hydrogen gas (H\(_2\)) is a valuable product in many industrial processes, including the synthesis of ammonia and hydrogenation reactions. The production of hydrogen gas from propane is an efficient method to supply hydrogen due to its high yield. Consider this balanced reaction:

\[ \text{C}_3\text{H}_8 + 4\text{H}_2\text{O} \rightarrow 3\text{CO} + 10\text{H}_2 \]

From the stoichiometric coefficients, we know that each mole of propane can produce ten moles of hydrogen gas. To calculate the mass of hydrogen gas produced, we first convert the amount of propane used to moles, then apply the stoichiometric ratio, and finally convert to mass using the molar mass of hydrogen.
  • Convert mass of propane to moles.
  • Use stoichiometry for mole conversion.
  • Convert moles of hydrogen to mass.

The significance of hydrogen production lies in its environmental and energy applications, making these calculations crucial for efficient and sustainable chemical production.

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Most popular questions from this chapter

When potassium cyanide ( \(\mathrm{KCN}\) ) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off. Here is the equation: $$ \mathrm{KCN}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{KCl}(a q)+\mathrm{HCN}(g) $$ If a sample of \(0.140 \mathrm{~g}\) of \(\mathrm{KCN}\) is treated with an excess of \(\mathrm{HCl}\), calculate the amount of HCN formed, in grams.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

Determine the empirical formulas of the compounds with the following compositions: (a) 40.1 percent \(\mathrm{C}\), 6.6 percent \(\mathrm{H}, 53.3\) percent \(\mathrm{O} ;\) (b) 18.4 percent \(\mathrm{C}\), 21.5 percent \(\mathrm{N}, 60.1\) percent \(\mathrm{K}\)

Suppose you are given a cube made of magnesium (Mg) metal of edge length \(1.0 \mathrm{~cm} .\) (a) Calculate the number of \(\mathrm{Mg}\) atoms in the cube. (b) Atoms are spherical in shape. Therefore, the \(\mathrm{Mg}\) atoms in the cube cannot fill all the available space. If only 74 percent of the space inside the cube is taken up by \(\mathrm{Mg}\) atoms, calculate the radius in picometers of an \(\mathrm{Mg}\) atom. (The density of \(\mathrm{Mg}\) is \(1.74 \mathrm{~g} / \mathrm{cm}^{3},\) and the volume of a sphere of radius \(r\) is \(\left.\frac{4}{3} \pi r^{3} .\right)\)

Platinum forms two different compounds with chlorine. One contains 26.7 percent \(\mathrm{Cl}\) by mass, and the other contains 42.1 percent \(\mathrm{Cl}\) by mass. Determine the empirical formulas of the two compounds.

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