Question

Coal gasification is a process that converts coal into methane gas. If this reaction has a percentage yield of \(85.0 \%,\) what mass of methane can be obtained from 1250 g of carbon? $$ 2 \mathrm{C}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}_{2}(g) $$

Short answer

708.328 g

Step-by-step solution

- Write the balanced chemical equation

The balanced chemical equation for the gasification process is given as: \[ 2 \mathrm{C}(s) + 2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{CH}_4(g) + \mathrm{CO}_2(g) \]

- Determine the molar masses

Calculate the molar masses of carbon (C) and methane (CH\textsubscript{4}): The atomic mass of carbon (C) is approximately 12 g/mol.The molar mass of methane (CH\textsubscript{4}) is calculated as:\[ 12 \, \mathrm{g/mol} \; \text{(C)} + 4 \times 1 \, \mathrm{g/mol} \; \text{(H)} = 16 \, \mathrm{g/mol} \; \text{(CH}_4) \]

- Convert mass of carbon to moles

Given 1250 g of carbon, convert this mass to moles of carbon: \[ n(\text{C}) = \frac{\text{mass}}{\text{molar mass}} = \frac{1250 \, \mathrm{g}}{12 \, \mathrm{g/mol}} = 104.167 \, \text{mol} \]

- Use mole ratio

According to the balanced chemical equation, 2 moles of carbon produce 1 mole of methane (CH\textsubscript{4}). Use this ratio to find the moles of methane: \[ n(\text{CH}_4) = \frac{1}{2} n(\text{C}) = \frac{1}{2} \times 104.167 = 52.083 \, \text{mol} \]

- Calculate theoretical yield of methane

Convert moles of methane to grams using the molar mass of methane: \[ \text{mass}(\text{CH}_4) = n(\text{CH}_4) \times \text{molar mass}(\text{CH}_4) = 52.083 \, \text{mol} \times 16 \, \mathrm{g/mol} = 833.328 \, \mathrm{g} \]

- Calculate actual yield of methane

Given the percentage yield is 85.0%, calculate the actual yield: \[ \text{Actual yield} = \text{percentage yield} \times \text{theoretical yield} = 0.85 \times 833.328 \, \mathrm{g} = 708.328 \, \mathrm{g} \]

Key concepts

Percentage Yield Calculation

To calculate the percentage yield, you need two main values: the actual yield and the theoretical yield. The theoretical yield is the amount of product expected from a chemical reaction, if everything goes perfectly. The actual yield is the amount actually produced.
The formula for percentage yield is:
\[ \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \] In this example, the theoretical yield of methane gas is 833.328 grams. Given a percentage yield of 85.0%, we can determine the actual yield as follows:
\[ \text{Actual Yield} = 0.85 \times 833.328 \; \text{g} = 708.328 \; \text{g} \] This means under given conditions, 708.328 grams of methane will be obtained.

Mole Ratio

In chemical reactions, the mole ratio is crucial for understanding how reactants convert into products. The balanced chemical equation gives you this ratio.
Consider our balanced equation:
\[ 2 \; \text{C} (s) + 2 \; \text{H}_2 \text{O} (l) \rightarrow \text{CH}_4 (g) + \text{CO}_2 (g) \] The coefficients (the numbers in front of the chemicals) tell you the mole ratio of the reactants and products. Here, 2 moles of carbon react with 2 moles of water to produce 1 mole of methane and 1 mole of carbon dioxide.
When we start with 1250 grams of carbon:

• Convert to moles: \[ n(\text{C}) = \frac{1250 \; \text{g}}{12 \; \text{g/mol}} = 104.167 \; \text{mol} \]
• Use the mole ratio to find moles of methane: \[ n(\text{CH}_4) = \frac{1}{2} \times 104.167 = 52.083 \; \text{mol} \]

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It helps us understand how much of each substance is involved.
The balanced chemical equation is the starting point. Here’s what we have:
\[ 2 \; \text{C} + 2 \; \text{H}_2 \text{O} \rightarrow \text{CH}_4 + \text{CO}_2 \] To solve for the amount of methane produced from a given amount of carbon, follow the steps:

• Calculate moles of each reactant.
• Use the mole ratio from the balanced equation.
• Convert moles of product to grams using molar mass.

For carbon: \[ n(\text{C}) = \frac{1250 \; \text{g}}{12 \; \text{g/mol}} = 104.167 \; \text{mol} \] For methane: \[ n(\text{CH}_4) = \frac{1}{2} \times 104.167 = 52.083 \; \text{mol} \] Then convert to mass: \[ \text{mass}(\text{CH}_4) = 52.083 \; \text{mol} \times 16 \; \text{g/mol} = 833.328 \; \text{g} \]

Theoretical Yield

Theoretical yield is the maximum amount of product that can be formed from the reactants. It assumes perfect conditions with no losses.
Based on the balanced chemical equation:
\[ 2 \; \text{C} + 2 \; \text{H}_2 \text{O} \rightarrow \text{CH}_4 + \text{CO}_2 \] To find the theoretical yield of methane:

• First, calculate moles of carbon: \[ n(\text{C}) = \frac{1250 \; \text{g}}{12 \; \text{g/mol}} = 104.17 \; \text{mol} \]
• Then, use the mole ratio to find moles of methane: \[ n(\text{CH}_4) = \frac{1}{2} \times 104.167 = 52.083 \; \text{mol} \]

Finally, convert to grams: \[ \text{mass}(\text{CH}_4) = 52.083 \; \text{mol} \times 16 \; \text{g/mol} = 833.328 \; \text{g} \]

Balanced Chemical Equation

A balanced chemical equation ensures that the law of conservation of mass is maintained. The number of atoms for each element is the same on both the reactant and product sides.
For coal gasification:
\[ 2 \; \text{C} (s) + 2 \; \text{H}_2 \text{O} (l) \rightarrow \text{CH}_4 (g) + \text{CO}_2 (g) \] This equation tells us:

• 2 atoms of carbon (C)
• 4 atoms of hydrogen (H) from water
• 2 atoms of oxygen (O)

On the product side, we get:

• 1 molecule of methane (\text{CH}_4) which has 1 atom of carbon (C) and 4 atoms of hydrogen (H)
• 1 molecule of carbon dioxide (\text{CO}_2) which has 1 atom of carbon (C) and 2 atoms of oxygen (O)

The equation is balanced, showing equal atoms of each element on both sides. This is essential for accurate stoichiometric calculations.