Problem 71 - Chemistry and Chemical Reactivity

Question

Iron carbonyl can be made by the direct reaction of iron metal and carbon monoxide. $$ \mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(\ell) $$ What is the theoretical yield of \(\mathrm{Fe}(\mathrm{CO})_{5}\), if \(3.52 \mathrm{g}\) of iron is treated with CO gas having a pressure of \(732 \mathrm{mm}\) Hg in a \(5.50-\) I. Hask at \(23^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

8.59 g of Fe(CO)_5 is the theoretical yield.

Step 1: Determine Moles of Iron

First, calculate the number of moles of iron (Fe). Iron has a molar mass of 55.85 g/mol. \[ \text{Moles of Fe} = \frac{3.52 \text{ g Fe}}{55.85 \text{ g/mol}} = 0.0630 \text{ moles Fe} \]

Step 2: Determine Moles of CO

Next, calculate the number of moles of CO using the ideal gas law equation: \[ PV = nRT \] Where:- \(P\) = pressure in atm = \(732 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 0.9632 \text{ atm} \)- \(V\) = volume in liters = 5.50 L- \(R\) = ideal gas constant = 0.0821 L·atm/mol·K- \(T\) = temperature in Kelvin = \(23^{\circ} \text{C} + 273.15 = 296.15 K\)Now calculate:\[ \text{Moles of CO} = \frac{PV}{RT} = \frac{0.9632 \times 5.50}{0.0821 \times 296.15} = 0.219 \text{ moles CO} \]

Step 3: Compare Moles to Find the Limiting Reactant

According to the balanced equation, 1 mole of Fe reacts with 5 moles of CO to produce 1 mole of Fe(CO)_5. - Moles of Fe available = 0.0630 - Moles of CO available = 0.219 Required ratio = 1:5 Check: - Fe: Needs 0.0630 × 5 = 0.315 moles of CO, but only 0.219 moles are available. Thus, CO is the limiting reactant.

Step 4: Calculate the Theoretical Yield in Moles of Fe(CO)_5

Since CO is the limiting reactant, use its moles to find out moles of Fe(CO)_5 formed.According to the reaction, 5 moles of CO form 1 mole of Fe(CO)_5, thus:\[ \text{Moles of Fe(CO)}_5 = \frac{0.219}{5} = 0.0438 \text{ moles Fe(CO)}_5 \]

Step 5: Calculate the Theoretical Yield in Grams of Fe(CO)_5

The molar mass of Fe(CO)_5 is calculated as follows: - Fe = 55.85 g/mol - CO = 12.01 + 16.00 = 28.01 g/mol- Fe(CO)_5 = 55.85 + 5(28.01) = 195.91 g/molUsing the moles from Step 4:\[ \text{Mass of Fe(CO)}_5 = 0.0438 \text{ moles} \times 195.91 \text{ g/mol} = 8.59 \text{ g} \]

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, limiting the amount of product that can be formed. Consider a scenario where you have two reactants, such as in our example of iron (\(\mathrm{Fe}\)) and carbon monoxide (\(\mathrm{CO}\)). These reactants combine to form iron pentacarbonyl (\(\mathrm{Fe(CO)}_5\)).- Consider the balanced equation: \(\mathrm{Fe} + 5 \mathrm{CO} \longrightarrow \mathrm{Fe(CO)}_{5} \)- Here, 1 mole of \(\mathrm{Fe}\) reacts with 5 moles of \(\mathrm{CO}\).- Calculate the moles of each reactant to identify which one will run out first. - In this case from our calculation, with 0.0630 moles of \(\mathrm{Fe}\) and 0.219 moles of \(\mathrm{CO}\), carbon monoxide (\(\mathrm{CO}\)) is the limiting reactant. Understanding the concept of the limiting reactant is key to calculate the amount of product that can actually be produced in a reaction, termed the theoretical yield.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and number of moles (\(n\)) of a gas. The equation is: \[ PV = nRT \]- \(R\) is the ideal gas constant, which is 0.0821 L·atm/mol·K.This law is crucial when dealing with gases, as it allows you to calculate one variable if the others are known. In the step-by-step solution, we used this law to find the number of moles of carbon monoxide (\(\mathrm{CO}\)).- Convert pressure from mm Hg to atm, as the ideal gas constant uses these units: \[ 732 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 0.9632 \text{ atm} \]- Volume is given as 5.50 L.- Temperature needs converting to Kelvin by adding 273.15: \[ 23^\circ \text{C} + 273.15 = 296.15 \text{ K} \]By substituting into the equation, you calculate moles of \(\mathrm{CO}\), helping to determine the limiting reactant accurately.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It involves balancing chemical equations and using these relationships for calculations.- Based on the balanced chemical equation for the reaction of iron with carbon monoxide: \(\mathrm{Fe} + 5\mathrm{CO} \rightarrow \mathrm{Fe(CO)}_{5} \)- The coefficients indicate the ratio of moles of reactants to products: 1 mole of \(\mathrm{Fe}\) to 5 moles of \(\mathrm{CO}\) produces 1 mole of\(\mathrm{Fe(CO)}_{5}\).Through stoichiometric calculations, we determined that \(\mathrm{CO}\) is the limiting reactant, hence defines the amount of product formed. This connection helps us calculate the actual amount of \(\mathrm{Fe(CO)}_{5}\) that can be formed from a given amount of the limiting reactant. Stoichiometry is a valuable tool in chemistry, allowing accurate predictions of product amounts in reactions.

Molar Mass Calculation

Molar mass is a measure of the mass of one mole of an element or compound, typically expressed in grams per mole (g/mol). Calculating molar mass is one of the fundamental skills in chemistry, essential for stoichiometry.To find the molar mass of iron pentacarbonyl (\(\mathrm{Fe(CO)}_{5}\)):- First, identify the molar mass of each component: - Iron (\(\mathrm{Fe}\)): 55.85 g/mol - Carbon monoxide (\(\mathrm{CO}\)): 12.01 (Carbon) + 16.00 (Oxygen) = 28.01 g/mol - Then calculate the molar mass for the entire compound: \[\text{Molar Mass of } \mathrm{Fe(CO)_{5}} = 55.85 + 5(28.01) = 195.91 \text{ g/mol}\]Understanding how to calculate molar mass allows you to convert between grams and moles, a necessary conversion step in many chemical calculations. In this example, it helped us determine the theoretical yield of\(\mathrm{Fe(CO)_{5}}\).