Problem 83 - Chemistry: An Atoms-Focused Approach

Question

Syngas An industrial process for producing hydrogen gas is based on the reaction between carbon heated to incandescence and steam that produces a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) called synthesis gas, or syngas. a. Write a balanced chemical equation for the production of syngas. b. If a reaction vessel that initially contains 66 kilograms of incandescent carbon and excess steam produces \(6.8 \mathrm{kg}\) \(\mathrm{H}_{2},\) what is the percent yield?

Step-by-Step Solution

Answer

Answer: The balanced chemical equation for the production of syngas is C + H2O + 1/2 O2 → CO + H2. The percent yield of hydrogen gas in the given reaction is 61.28%.

Step 1: Write the balanced chemical equation

Start by writing the unbalanced chemical equation between carbon (C) and steam (\(\mathrm{H}_2\mathrm{O}\)) to produce syngas (\(\mathrm{CO}\) and \(\mathrm{H}_2\)): C + H\(_2\)O \(\rightarrow\) CO + H\(_2\) Then, balance the equation: C + H\(_2\)O \(\rightarrow\) CO + H\(_2\) We need an additional Oxygen atom on the left side of the reaction to balance the equation. Consequently, we add 1/2 O\(_2\) to the reactants: C + H\(_2\)O + 1/2 O\(_2\) \(\rightarrow\) CO + H\(_2\) Now, the equation should be balanced.

Step 2: Calculate moles of carbon and hydrogen

We are given the following information: - 66 kg of carbon (m_C = 66 kg) - 6.8 kg of hydrogen gas produced (m_H2 = 6.8 kg) First, we need to convert the masses into moles using the formula: \(n=\frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass. For carbon: Molar mass of carbon (C) = 12.01 g/mol Convert the mass of carbon to grams: m_C = 66 kg × 1000 g/kg = 66000 g n_C = \(\frac{m_C}{M_C}\) = \(\frac{66000\text{ g}}{12.01\text{ g/mol}}=5495.4\text{ mol}\) For hydrogen gas: Molar mass of hydrogen gas (H\(_2\)) = 2.02 g/mol Convert the mass of hydrogen gas to grams: m_H2 = 6.8 kg × 1000 g/kg = 6800 g n_H2 = \(\frac{m_H2}{M_H}\)= \(\frac{6800\text{ g}}{2.02\text{ g/mol}}=3366.3\text{ mol}\)

Step 3: Calculate theoretical yield of hydrogen gas

From the balanced chemical equation: C + H\(_2\)O + 1/2 O\(_2\) \(\rightarrow\) CO + H\(_2\) We can see that 1 mole of carbon (C) produces 1 mole of hydrogen gas (H\(_2\)). Thus, the theoretical yield of H\(_2\) (in moles) should be equal to the moles of carbon (C) in the reaction. Theoretical yield of H\(_2\) (mol) = 5495.4 mol

Step 4: Calculate percent yield

Now we will calculate the percent yield using the formula: Percent yield = \(\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%\) The actual yield of H\(_2\) (mol) = 3366.3 mol Theoretical yield of H\(_2\) (mol) = 5495.4 mol Percent Yield = \(\frac{3366.3\text{ mol}}{5495.4\text{ mol}}\times100\%=61.28\%\) The percent yield of hydrogen gas in this reaction is 61.28%.

Key Concepts

Balanced Chemical Equation

When forming syngas from a carbon and steam reaction, it's crucial to represent the process with a balanced chemical equation. Initially, the reaction can be stated as: \( \text{C} + \text{H}_2\text{O} \rightarrow \text{CO} + \text{H}_2 \). To balance it correctly, ensure the number of each type of atom is identical on both sides.
Notice that, as written, there is an insufficient amount of oxygen atoms on the left compared to the right. To balance them, add \( \frac{1}{2} \text{O}_2 \) on the reactant side, resulting in the balanced equation: \( \text{C} + \text{H}_2\text{O} + \frac{1}{2} \text{O}_2 \rightarrow \text{CO} + \text{H}_2 \).
A balanced chemical equation is vital as it reflects the conservation of mass and atoms during a reaction. Keeping track of all reactants and products helps in subsequent calculations like finding theoretical yields.

Molar Mass Calculation

Understanding molar mass is key to linking mass and the number of moles, crucial for solving chemistry problems. Molar mass is the mass of one mole of a substance, expressed in grams per mole \( \text{g/mol} \).
For carbon, the molar mass is 12.01 \( \text{g/mol} \). This means every mole of carbon weighs 12.01 grams. Given 66 kilograms of carbon, you first convert this to grams (66,000 g) and calculate the moles: \( n = \frac{66000}{12.01} = 5495.4 \text{ mol} \).
For hydrogen gas \( \text{H}_2 \), with a molar mass of 2.02 \( \text{g/mol} \), convert 6.8 kg to 6800 g, then compute the moles: \( n = \frac{6800}{2.02} = 3366.3 \text{ mol} \).
These calculations help in determining how much of each reactant is used and the amount of product formed, fundamental for finding theoretical and percent yield.

Percent Yield

Percent yield measures the efficiency of a chemical reaction. It compares the actual yield (what you obtained) to the theoretical yield (what you expected, based on stoichiometry). This is expressed as a percentage using the formula:

• Percent Yield = \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \).

In the example, the actual yield of hydrogen gas is 3366.3 mol, whereas the theoretical yield is 5495.4 mol. The calculated percent yield is thus \( \frac{3366.3}{5495.4} \times 100\% = 61.28\% \).
A percent yield less than 100% may indicate incomplete reactions or loss of products during the process. Understanding why the yield isn't 100% can guide process improvements and increase efficiency.

Theoretical Yield

Theoretical yield represents the maximum amount of product that can be obtained from a reaction, assuming complete conversion of reactants to products. To determine this, use stoichiometry involving the balanced chemical equation.
For the syngas reaction, since 1 mole of carbon reacts to produce 1 mole of hydrogen gas \( \text{H}_2 \), the theoretical yield of hydrogen gas directly equals the moles of carbon used.
Given 5495.4 mol of carbon, the theoretical yield of \( \text{H}_2 \) is similarly 5495.4 mol. This calculation uses the premise that the reaction goes to completion without losses.
Understanding the theoretical yield is essential for assessing the practical efficiency of a chemical process and for calculation of the percent yield.