Problem 131 - Chemistry The Science in Context

Question

The formation of organic compounds by the reaction of iron(II) sulfide with carbonic acid is described by the following chemical equation:$$2 \mathrm{FeS}+\mathrm{H}_{2} \mathrm{CO}_{3} \rightarrow 2 \mathrm{FeO}+1 / n\left(\mathrm{CH}_{2} \mathrm{O}\right)_{n}+2 \mathrm{S}$$ a. How much FeO is produced starting with \(1.50 \mathrm{g}\) of FeS and 0.525 mol of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) if the reaction results in a \(78.5 \%\) yield? "b. If the carbon-containing product has a molar mass of \(3.00 \times 10^{2} \mathrm{g} / \mathrm{mol},\) what is the chemical formula of the product?

Step-by-Step Solution

Answer

Answer: The mass of FeO produced is approximately 0.958 g, and the chemical formula of the carbon-containing product is (CH2O)_{10}.

Step 1: Calculate the moles of FeS

To calculate the moles of FeS, use the formula: moles of FeS = mass of FeS / molar mass of FeS The molar mass of FeS is approximately 56 g/mol (Fe) + 32 g/mol (S) = 88 g/mol. Therefore, the moles of FeS are: moles = 1.50 g / 88 g/mol ≈ 0.017 mol

Step 2: Determine the limiting reactant

Since the reaction equation is balanced, we can compare the moles of FeS and H2CO3 to determine the limiting reactant. From the equation, we see that 2 moles of FeS react with 1 mole of H2CO3. Therefore, we need to compare the moles of H2CO3 to the moles of FeS by dividing by their coefficients. The smallest ratio will determine the limiting reactant. FeS ratio = 0.017 mol / 2 H2CO3 ratio = 0.525 mol / 1 The FeS ratio is 0.0085, whereas the H2CO3 ratio is 0.525, so FeS is the limiting reactant.

Step 3: Calculate the theoretical yield of FeO

The stoichiometry of the reaction shows that 2 moles of FeS produce 2 moles of FeO. Since FeS is the limiting reactant, we can use its moles to find the moles of FeO produced. moles of FeO = moles of FeS = 0.017 mol

Step 4: Calculate the actual yield of FeO

The reaction results in a 78.5% yield of FeO, so the actual yield can be calculated as: actual moles of FeO = moles of FeO × percentage yield actual moles of FeO = 0.017 × 0.785 ≈ 0.0133 mol

Step 5: Calculate the mass of FeO produced

To find the mass of FeO produced, use the formula: mass of FeO = moles of FeO × molar mass of FeO The molar mass of FeO is approximately 56 g/mol (Fe) + 16 g/mol (O) = 72 g/mol. Therefore, the mass of FeO produced is: mass = 0.0133 mol × 72 g/mol ≈ 0.958 g #b. Finding the chemical formula of the carbon-containing product#

Step 1: Determine the molar mass of CH2O

The molar mass of CH2O is approximately 12 g/mol (C) + 2 g/mol (H) + 16 g/mol (O) = 30 g/mol.

Step 2: Calculate the value of n

To find the value of n, we need to divide the given molar mass of the carbon-containing product by the molar mass of CH2O: n = molar mass of the product / molar mass of CH2O n = 3.00 × 10^2 g/mol / 30 g/mol = 10

Step 3: Write the chemical formula

Now that we know the value of n, we can write the chemical formula of the carbon-containing product: (CH2O)_{10}

Key Concepts

Limiting Reactant

In a chemical reaction, determining the limiting reactant is crucial to understanding how much product will be formed. The limiting reactant is the substance that gets completely consumed first, limiting the extent of the reaction.
Let's take a look at how you might identify it in a scenario where iron(II) sulfide (FeS) reacts with carbonic acid (H₂CO₃).
Given the equation: \[2 \, \mathrm{FeS} + \mathrm{H}_{2} \mathrm{CO}_{3} \rightarrow 2 \, \mathrm{FeO} + ...\] You'll first need to determine the number of moles of each reactant:

• For FeS, convert grams to moles: \(1.50 \, \text{g} \div 88 \, \mathrm{g/mol} \approx 0.017 \, \text{mol}\).
• For H₂CO₃, it is given as 0.525 mol.

Now compare these moles to their coefficients in the balanced equation:

• \(\text{Ratio for FeS: } 0.017 \, \text{mol} \div 2 = 0.0085\)
• \(\text{Ratio for H}_{2}\text{CO}_{3}: 0.525 \, \text{mol} \div 1 = 0.525\)

The smaller ratio indicates the limiting reactant. Here, FeS is the limiting reactant because it has the smallest ratio.

Theoretical Yield

Theoretical yield represents the maximum amount of product that can be generated from a given amount of reactants, assuming perfect conditions and complete conversion.
It's calculated based on the stoichiometry of the balanced chemical equation.
For the reaction between FeS and H₂CO₃, you use the balanced equation to find the amount of iron oxide (FeO) expected from FeS:\[\text{Starting with 0.017 mol of FeS: } 2 \, \text{mol FeS} \rightarrow 2 \, \text{mol FeO}\] Since 2 moles of FeS produce 2 moles of FeO, the theoretical yield in moles for FeO matches the moles of FeS.With FeS as the limiting reactant, the moles of FeO are:

• \(0.017 \, \text{mol of FeO}\).

Then convert this to grams if needed, using the molar mass of FeO (72 g/mol).\[\text{Mass of FeO} = 0.017 \, \text{mol} \times 72 \, \text{g/mol} \approx 1.22 \, \text{g}\]

Percent Yield

In reality, reactions rarely proceed to complete conversion of reactants into products. Percent yield helps us understand the efficiency of a reaction.
It's calculated by comparing the actual yield (the amount of product obtained in practice) with the theoretical yield.
The formula for percent yield is:\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]For our reaction:

• The theoretical yield of FeO was 1.22 g (from earlier calculation).
• The actual yield was calculated as 0.958 g using the yield percentage (78.5%).

Substitute these values into the formula:\[\text{Percent Yield for FeO} = \left( \frac{0.958 \, \text{g}}{1.22 \, \text{g}} \right) \times 100\% = 78.5\%\]This means only 78.5% of the theoretical amount was produced due to various practical limitations.