Problem 157 - Chemistry The Molecular Science

Question

Dicobalt octacarbonyl, \(\mathrm{Co}_{2}(\mathrm{CO})_{8},\) is formed by the reaction of cobalt metal with carbon monoxide gas. (a) Calculate the maximum mass of dicobalt octacarbonyl that can be formed from \(100.0 \mathrm{~g}\) cobalt and \(200.0 \mathrm{~g}\) carbon monoxide, assuming \(89.7 \%\) yield. (b) Assuming that the deviation from \(100 \%\) yield was entirely due to loss of product during purification, calculate the additional mass of the limiting reactant that would be needed to react completely with the residual excess reactant. (c) Calculate the total mass of dicobalt octacarbonyl produced in the reactions in parts (a) and (b).

Step-by-Step Solution

Answer

273.84 g of dicobalt octacarbonyl was produced.

Step 1: Identify the Reactants and the Product

The reactants are cobalt metal (Co) and carbon monoxide gas (CO), and the product is dicobalt octacarbonyl, Co2(CO)8.

Step 2: Calculate Molar Masses

The molar mass of Co is 58.93 g/mol and CO is 28.01 g/mol. The molar mass of Co2(CO)8 is calculated as \(2 \times 58.93 + 8 \times 28.01 = 341.92\) g/mol.

Step 3: Write the Balanced Chemical Equation

The balanced chemical equation for this reaction is: \[ 2 \text{Co} + 8 \text{CO} \rightarrow \text{Co}_2(\text{CO})_8 \]

Step 4: Determine the Limiting Reactant

Convert the given masses of Co and CO to moles. \[ \text{Moles of Co} = \frac{100.0}{58.93} \approx 1.697 \text{ moles} \] and \[ \text{Moles of CO} = \frac{200.0}{28.01} \approx 7.142 \text{ moles} \]. The stoichiometry requires 8 moles of CO for every 2 moles of Co. So, the limiting reactant is Co.

Step 5: Calculate Maximum Mass of Product in 100% Yield

Since Co is the limiting reactant, use its moles to find the theoretical yield. The stoichiometry indicates that 1 mole of Co2(CO)8 is formed from 2 moles of Co. \( \text{Moles of Co}_2(\text{CO})_8 = \frac{1.697}{2} = 0.8485 \text{ moles} \). Calculate the mass: \(0.8485 \times 341.92 = 290.10 \text{ g} \).

Step 6: Adjust for the Percent Yield

Since the yield is 89.7%, the actual mass obtained is: \(290.10 \times 0.897 = 260.27 \text{ g} \).

Step 7: Calculate Unused Reactant

Calculate how much CO is used up with the limiting reactant (Co). Moles of CO used: \(1.697 \times \frac{8}{2} = 6.788 \text{ moles} \). Unused CO: \(7.142 - 6.788 = 0.354 \text{ moles} \), or \(0.354 \times 28.01 = 9.92 \text{ g} \) of CO.

Step 8: Determine Additional Limiting Reactant Required

To use the remaining 9.92 g of CO fully, convert to moles \((\frac{9.92}{28.01} = 0.354 \text{ moles})\). The stoichiometry requires \(\frac{0.354}{8} \times 2 = 0.0885 \text{ moles of Co}\). Convert to mass: \(0.0885 \times 58.93 = 5.21 \text{ g}\).

Step 9: Calculate Additional Product Formed in 100% Yield

Using the additional Co, convert moles to grams: \((\frac{0.0885}{2} = 0.04425 \, \text{moles of } \text{Co}_2(\text{CO})_8) \to 0.04425 \times 341.92 = 15.13 \text{ g}\) in 100% yield. Adjust for yield: \(15.13 \times 0.897 = 13.57 \text{ g}\).

Step 10: Calculate Total Mass Produced

Sum the masses from both reactions: \(260.27 \text{ g} + 13.57 \text{ g} = 273.84 \text{ g}\).

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is entirely consumed first during the reaction, determining the amount of product formed. To identify it, we compare the actual mole ratio of reactants used with the stoichiometric ratio from the balanced equation. Given our equation:\[ 2 \text{Co} + 8 \text{CO} \rightarrow \text{Co}_2(\text{CO})_8 \]The limiting reactant can be identified by converting the masses of the reactants to moles. For cobalt (Co), using its molar mass 58.93 g/mol, the moles are:\[ \text{Moles of Co} = \frac{100.0}{58.93} \approx 1.697 \text{ moles} \]For carbon monoxide (CO), with a molar mass of 28.01 g/mol, the moles are:\[ \text{Moles of CO} = \frac{200.0}{28.01} \approx 7.142 \text{ moles} \]The balanced equation requires 8 moles of CO for every 2 moles of Co. Comparing the required and available moles determines Co as the limiting reactant, since we have less Co than needed per the stoichiometric calculations.

Percent Yield

Percent yield is used to determine the efficiency of a reaction by comparing the actual yield to the theoretical yield. It’s calculated using the formula:\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]In the problem scenario, the theoretical yield is calculated using the limiting reactant. With cobalt as the limiting reactant, it allows for theoretically 0.8485 moles of dicobalt octacarbonyl to be formed, giving a mass of:\[ 0.8485 \times 341.92 = 290.10 \text{ g} \]But, the actual yield is 89.7% of this due to losses during purification:\[ \text{Actual Yield} = 290.10 \times 0.897 = 260.27 \text{ g} \]This adjustment shows how practical lab results can differ due to imperfect conditions.

Chemical Equation Balancing

Balancing chemical equations ensures that the number of atoms for each element is equal on both sides of the equation, reflecting the conservation of mass. Our equation is:\[ 2 \text{Co} + 8 \text{CO} \rightarrow \text{Co}_2(\text{CO})_8 \]Here, the balancing aligns with the stoichiometry:

• 2 Co atoms on the left react to form 1 Co2 entity on the right.
• 8 CO molecules on the left lead to the same number in form of octacarbonyl groups in the product on the right.

Balancing ensures each element is accounted for mass-wise, vital for both practical and theoretical chemistry applications.

Molar Mass Calculation

Molar mass calculation is critical for converting between mass and moles in stoichiometry. It reflects the mass of one mole of a substance, measured in g/mol. In the given exercise:

• Molar mass of Co is calculated as 58.93 g/mol.
• Molar mass of CO is calculated as 28.01 g/mol.
• Molar mass of dicobalt octacarbonyl, Co2(CO)8, is calculated through its composition: \( 2 \times 58.93 + 8 \times 28.01 = 341.92 \text{ g/mol} \).

This calculation helps us convert given masses of Co and CO into moles, which are used to determine the limiting reactant and eventual product yields. Mastering molar mass calculations is a cornerstone of chemical problem-solving and reactions analysis.