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Chapter 5: Problem 165

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). In a certain experiment, \(20.00 \mathrm{g}\) iron metal was reacted with \(11.20 \mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 g oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

Short Answer

Expert verified
In the given experiment, 46.84 g of FeO and 20.76 g of Fe2O3 were produced.

Step by step solution

01

Determine the moles of Fe and O2

First, we need to find the moles of Fe and O2 involved in the reaction. To do this, we will divide the given masses by the respective molecular weights. For Fe: Molecular weight = 55.85 g/mol Moles = (20.00 g) / (55.85 g/mol) = 0.358 mol Fe For O2: Molecular weight = 32.00 g/mol Moles = (11.20 g) / (32.00 g/mol) = 0.350 mol O2
02

Determine the maximum possible amounts of FeO and Fe2O3 formed

Now that we have the moles of Fe and O2, let's consider the stoichiometry of the two possible reactions: 1. Fe + O2 -> 2 FeO 2. 4 Fe + 3 O2 -> 2 Fe2O3 Case 1: If all Fe reacts to form FeO: Moles of FeO = 2*(moles of Fe) = 2*0.358 = 0.716 mol Case 2: If all Fe reacts to form Fe2O3: Moles of Fe2O3 = 0.5*(moles of Fe) = 0.5*0.358 = 0.179 mol We have to determine which one is the limiting reactant.
03

Identify the limiting reactant

Let's compare the amount of O2 needed for both reactions: Case 1: For 0.716 mol of FeO, the required O2 = 0.716/2 = 0.358 mol Case 2: For 0.179 mol of Fe2O3, the required O2 = (3/2)*0.179 = 0.2685 mol We have 0.350 mol of O2 available. Since both reactions would require less O2 than what is available, the actual reaction would be a mixture of both reaction 1 and 2. To account for this, we need to find out how much FeO and Fe2O3 are formed when all the iron is consumed.
04

Determine the amounts of FeO and Fe2O3 formed

In the given problem, we have a remaining mass of 3.24 g of oxygen after the reaction. This means that 11.20 g - 3.24 g = 7.96 g of oxygen reacted. Now, let's label the moles of FeO formed as x and the moles of Fe2O3 formed as y. We need to solve the following two equations using this information: 1) (1/2)*x + (4/3)*y = 0.358 (from iron moles conservation) 2) x + (3/2)*y = (7.96 g) * (1/32.00 g/mol) = 0.249 mol (from oxygen moles conservation) By solving these two equations for x and y, we get: x = 0.652 mol y = 0.130 mol
05

Calculate the masses of FeO and Fe2O3 formed

Now that we have the moles of FeO and Fe2O3 formed after the reaction, we can calculate their respective masses. For FeO: Molecular weight = (55.85 g/mol) + (16.00 g/mol) = 71.85 g/mol Mass = (0.652 mol) * (71.85 g/mol) = 46.84 g For Fe2O3: Molecular weight = (2*55.85 g/mol) + (3*16.00 g/mol) = 159.69 g/mol Mass = (0.130 mol) * (159.69 g/mol) = 20.76 g In conclusion, the experiment produced 46.84 g of FeO and 20.76 g of Fe2O3.

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Most popular questions from this chapter

An element X forms both a dichloride (XCI_) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right) .\) Treatment of \(10.00 \mathrm{g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{g} \mathrm{XCl}_{4} .\) Calculate the atomic mass of \(\mathrm{X},\) and identify \(\mathrm{X}\).

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. \(\mathrm{A}\) \(752-\mathrm{g}\) sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{g}\) of pure iron by the following reaction: $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g)$$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. \(\operatorname{SiO}_{2}(s)+\mathrm{C}(s) \frac{\text { Electric }}{\text { arc furnace }} \mathrm{Si}(s)+\mathrm{CO}(g).\) b. Liquid silicon tetrachloride is reacted with very pure solid magnesium, producing solid silicon and solid magnesium chloride. c. \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+\mathrm{NaF}(s).\)

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right) .\) Calculate the percent composition by mass of fluoxymesterone.

Aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is synthesized by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right) .\) The balanced equation is $$\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$$ a. What mass of acetic anhydride is needed to completely consume \(1.00 \times 10^{2}\) g salicylic acid? b. What is the maximum mass of aspirin (the theoretical yield) that could be produced in this reaction?
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