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Iron carbonyl can be made by the direct reaction of iron metal and carbon monoxide. $$ \mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Fe}(\mathrm{CO})_{5}(\ell) $$ 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 \(732 \mathrm{mm} \mathrm{Hg}\) in a \(5.50-\mathrm{L}\). flask at \(23^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The theoretical yield is 8.54 g of Fe(CO)_5.

Step by step solution

01

Calculate Moles of Iron

First, calculate the number of moles of iron. The molar mass of iron (Fe) is \(55.85\ \text{g/mol}\). Use the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).\[moles\ \text{ of } Fe = \frac{3.52\ \text{g}}{55.85\ \text{g/mol}} = 0.063\ \text{mol}\]So, the moles of iron is \(0.063\ \text{mol}\).
02

Calculate Moles of CO

Use the ideal gas law to calculate the moles of CO gas. The ideal gas law is \(PV = nRT\), where \(P\) is pressure in atm, \(V\) is volume in L, \(n\) is moles, \(R\) is the gas constant \(0.0821\ \text{L atm/mol K}\), and \(T\) is the temperature in Kelvin.First, convert pressure from mm Hg to atm:\[P = \frac{732\ \text{mm Hg}}{760\ \text{mm Hg/atm}} = 0.963\ \text{atm}\]Next, convert temperature to Kelvin:\[T = 23 + 273.15 = 296.15\ \text{K}\]Now, solve for \(n\):\[0.963\ \text{atm} \times 5.50\ \text{L} = n \times 0.0821\ \text{L atm/mol K} \times 296.15\ \text{K}\]\[n = \frac{0.963 \times 5.50}{0.0821 \times 296.15} \approx 0.218\ \text{mol CO}\]
03

Determine Limiting Reactant

In the reaction, 1 mole of Fe requires 5 moles of CO. Compare the available mole ratio:For Fe:\(0.063\ \text{mol Fe} \rightarrow 0.315\ \text{mol CO needed (0.063 mol} \times 5)\)For CO:\(0.218\ \text{mol CO available}\)Since 0.315 mol of CO is needed, but only 0.218 mol is available, CO is the limiting reactant.
04

Calculate Theoretical Yield of Fe(CO)_5

According to the reaction, 5 moles of CO yield 1 mole of \(\text{Fe(CO)}_5\). As CO is the limiting reactant, use its moles to find the yield:\[\text{Moles of } \text{Fe(CO)}_5 = \frac{0.218\ \text{mol CO}}{5} = 0.0436\ \text{mol Fe(CO)}_5\]The molar mass of \(\text{Fe(CO)}_5\) is \(195.9\ \text{g/mol}\), so:\[\text{Mass of } \text{Fe(CO)}_5 = 0.0436\ \text{mol} \times 195.9\ \text{g/mol} = 8.54\ \text{g}\]Therefore, the theoretical yield of \(\text{Fe(CO)}_5\) is \(8.54\ \text{g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limiting Reactant
When it comes to chemical reactions, the concept of the limiting reactant is crucial. This is because it determines the maximum amount of product that can be formed. The limiting reactant is the substance that is completely consumed when the reaction goes to completion. In our example, we have iron (Fe) and carbon monoxide (CO) reacting to form iron carbonyl. To find out which reactant is limiting, we must first calculate the amount of each reactant available and see which one runs out first. This involves comparing the mole ratios. For our reaction, 1 mole of iron requires 5 moles of carbon monoxide to complete the reaction as per the equation: Fe + 5 CO → Fe(CO)₅. In the given problem, the moles of iron are 0.063, which would require 0.315 moles of CO. However, we only have 0.218 moles of CO available, making carbon monoxide the limiting reactant. Since there is not enough CO to convert all of the available iron into product, CO limits the reaction and thus determines the theoretical yield of iron carbonyl.
Ideal Gas Law
The ideal gas law is an essential tool for connecting the physical properties of gases: pressure (P), volume (V), and temperature (T) with the amount of gas in moles (n). It is expressed as:\(PV = nRT\).In this equation, \(R\) is the ideal gas constant, which equals 0.0821 L atm/mol K. To solve for the number of moles in a gas sample, rearrange the equation to: \(n = \frac{PV}{RT}\).Let's apply this to find the moles of CO in our reaction. We start by converting the pressure from mm Hg to atm, as calculations using the ideal gas law require standard units.1 atm = 760 mm Hg, so \(732 \text{ mm Hg} = 0.963 \text{ atm}\).Also, the temperature needs to be in Kelvin:\(T(\text{K}) = 23 + 273.15 = 296.15 \text{ K}\).Now we can find the moles of CO:\(n = \frac{(0.963 \text{ atm})(5.50 \text{ L})}{(0.0821 \text{ L atm/mol K})(296.15 \text{ K})} \approx 0.218 \text{ mole CO}\).This calculation correctly shows that the amount of CO available for the reaction is 0.218 moles, another hint towards identifying the limiting reactant.
Molar Mass Calculation
To understand productivity in chemical reactions, knowing how to calculate molar mass is essential. Molar mass, often measured in grams per mole, is important for converting between the mass of a substance and its moles.To calculate it:- Add up the atomic masses of all elements in the compound (using the periodic table for atomic masses).For instance, iron carbonyl \(\text{Fe(CO)}_5\) requires knowledge of atomic masses of iron and carbon monoxide. The molar mass of iron is 55.85 g/mol, and carbon (12.01 g/mol) plus oxygen (16.00 g/mol) gives CO a molar mass of 28.01 g/mol. Therefore, for \(\text{Fe(CO)}_5\):- Calculate as: \(55.85 \text{ g/mol for Fe} + 5 \times 28.01 \text{ g/mol for CO} = 195.9 \text{ g/mol}\).This molar mass helps determine both the moles of \(\text{Fe(CO)}_5\) possible from limiting CO and ultimately compute the theoretical yield (i.e., predicted mass of product resulting from a complete reaction of limiting reactant). In our example case, with 0.0436 moles of \(\text{Fe(CO)}_5\) produced and a molar mass of 195.9 g/mol, the theoretical yield is 8.54 grams.

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Most popular questions from this chapter

Chloroform is a common liquid used in the laboratory. It vaporizes readily. If the pressure of chloroform vapori in a flask is \(195 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C}\) and the density of the vapor is \(1.25 \mathrm{g} / \mathrm{L},\) what is the molar mass of chloroform?

You have a gas, one of the three known phosphorusfluorine compounds \(\left(\mathrm{PF}_{3}, \mathrm{PF}_{5}, \text { and } \mathrm{P}_{2} \mathrm{F}_{4}\right) .\) To find out which, you have decided to measure its molar mass. (a) First, you determine that the density of the gas is \(5.60 \mathrm{g} / \mathrm{L}\) at a pressure of \(0.971 \mathrm{atm}\) and a temperature of \(18.2^{\circ} \mathrm{C} .\) Calculate the molar mass and identify the compound. (b) To check the results from part (a), you decide to measure the molar mass based on the relative rates of effusion of the unknown gas and \(\mathrm{CO}_{2} .\) You find that \(\mathrm{CO}_{2}\) effuses at a rate of \(0.050 \mathrm{mol} / \mathrm{min}\) whereas the unknown phosphorus fluoride effuses at a rate of \(0.028 \mathrm{mol} / \mathrm{min.}\) Calculate the molar mass of the unknown gas based on these results.

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of \(\mathrm{O}_{2}\) (L) is required for complete reaction with \(5.2 \mathrm{L}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\) What volume of \(\mathrm{H}_{2} \mathrm{O}\) vapor ( \(\mathrm{L}\) ) is produced? Assume all gases are measured at the same temperature and pressure.

\(\mathrm{Ni}(\mathrm{CO})_{4}\) can be made by reacting finely divided nickel with gaseous CO. If you have CO in a \(1.50-\mathrm{L}\). flask at a pressure of \(418 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C},\) along with \(0.450 \mathrm{g}\) of Ni powder, what is the theoretical yield of \(\mathrm{Ni}(\mathrm{CO})_{4} ?\)

Silane, \(\operatorname{SiH}_{4},\) reacts with \(\mathrm{O}_{2}\) to give silicon dioxide and water: $$\mathrm{SiH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ A \(5.20-\) L sample of \(\mathrm{SiH}_{4}\) gas at \(356 \mathrm{mm}\) Hg pressure and \(25^{\circ} \mathrm{C}\) is allowed to react with \(\mathrm{O}_{2}\) gas. What volume of O, gas, in liters, is required for complete reaction if the oxygen has a pressure of \(425 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C} ?\)

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