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Chapter 3: Problem 153

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 \(3.24 \mathrm{~g}\) 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 this experiment, 12.86 g of FeO and 7.46 g of Fe鈧侽鈧� were formed.

Step by step solution

01

Write the balanced chemical equations for both reactions.

We need to consider the two reactions between iron and oxygen separately. The balanced chemical equations for these reactions are: 1. For the synthesis of FeO: \(2 \: Fe + O_2 \rightarrow 2 \: FeO\) 2. For the synthesis of Fe鈧侽鈧�: \(4 \: Fe + 3 \: O_2 \rightarrow 2 \: Fe_2O_3\)
02

Determine the moles of reactants.

Calculate the moles of iron and oxygen present initially using their molar masses. Here, molar mass of Fe is 55.85 g/mol and molar mass of O鈧� is 32.00 g/mol. Moles of Fe = \( \frac{20.00 \: g}{55.85 \: g/mol} = 0.358 \: mol\) Moles of O鈧� = \( \frac{11.20 \: g}{32.00 \: g/mol} = 0.350 \: mol\)
03

Calculate the moles of oxygen reacted.

Subtract the moles of remaining oxygen gas from moles of initial oxygen to determine the moles of reacted oxygen: Moles of O鈧� reacted = Initial moles of O鈧� - Remaining moles of O鈧� Remaining moles of O鈧� = \( \frac{3.24 \: g}{32.00 \: g/mol} = 0.101 \: mol\) Moles of O鈧� reacted = 0.350 mol - 0.101 mol = 0.249 mol
04

Determine moles of each compound formed.

Use the stoichiometry of the reactions to determine the moles of FeO and Fe鈧侽鈧� produced. As 0.358 mol of Fe reacted and no Fe was left, the maximum moles of FeO and Fe鈧侽鈧� formed can be: Maximum moles of FeO = \( \frac{0.358 \: mol \: Fe}{2} = 0.179 \: mol\) Maximum moles of Fe鈧侽鈧� = \( \frac{0.358 \: mol \: Fe}{4} = 0.0895 \: mol\) Using these values and the stoichiometry of equation 1, we can find the amount of oxygen used for FeO formation. Moles of O鈧� used for FeO formation = \(0.179 \: mol\) Since we have 0.249 mol of O鈧� reacted, the remaining oxygen must be consumed for Fe鈧侽鈧� formation. Moles of O鈧� used for Fe鈧侽鈧� formation = Moles of O鈧� reacted - Moles of O鈧� used for FeO formation = 0.249 mol - 0.179 mol = 0.070 mol Now, using the stoichiometry of equation 2, calculate the moles of Fe鈧侽鈧� formed: Moles of Fe鈧侽鈧� formed = \( \frac{0.070 \: mol \: O鈧倉{3} \times 2 = 0.0467 \: mol\)
05

Calculate the mass of each compound formed.

Use the molar masses of FeO and Fe鈧侽鈧� to calculate the mass of each compound formed: Mass of FeO formed = Moles of FeO 脳 Molar mass of FeO = 0.179 mol 脳 (55.85 g/mol + 16.00 g/mol) = 0.179 mol 脳 71.85 g/mol = 12.86 g Mass of Fe鈧侽鈧� formed = Moles of Fe鈧侽鈧� 脳 Molar mass of Fe鈧侽鈧� = 0.0467 mol 脳 (2 脳 55.85 g/mol + 3 脳 16.00 g/mol) = 0.0467 mol 脳 159.70 g/mol = 7.46 g
06

Answer

In this experiment, 12.86 g of FeO and 7.46 g of Fe鈧侽鈧� were formed.

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Most popular questions from this chapter

A \(2.077-g\) sample of an element, which has an atomic mass between 40 and 55 , reacts with oxygen to form \(3.708 \mathrm{~g}\) of an oxide. Determine the formula of the oxide (and identify the element).

Consider the following balanced chemical equation: $$ A+5 B \longrightarrow 3 C+4 D $$ a. Equal masses of \(\mathrm{A}\) and \(\mathrm{B}\) are reacted. Complete each of the following with either "A is the limiting reactant because \(" ;{ }^{\prime *} \mathrm{~B}\) is the limiting reactant because \("\) or "we cannot determine the limiting reactant because i. If the molar mass of \(\mathrm{A}\) is greater than the molar mass of B, then ii. If the molar mass of \(\mathrm{B}\) is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound \(\mathrm{A}\) has the same molar mass as carbon dioxide. Compound \(\mathrm{B}\) is a diatomic molecule. Identify \(\mathrm{com}-\) pound \(\mathrm{B}\) and support your answer. c. Compound \(\mathrm{A}\) is a hydrocarbon that is \(81.71 \%\) carbon by mass. Detemine its empirical and molecular formulas.

In using a mass spectrometer, a chemist sees a peak at a mass of \(30.0106\). Of the choices \({ }^{12} \mathrm{C}_{2}{ }^{1} \mathrm{H}_{6},{ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}\), and \({ }^{14} \mathrm{~N}^{16} \mathrm{O}\), which is responsible for this peak? Pertinent masses are \({ }^{1} \mathrm{H}, 1.007825\); \({ }^{16} \mathrm{O}, 15.994915 ;\) and \({ }^{14} \mathrm{~N}, 14.003074 .\)

The empirical formula of styrene is \(\mathrm{CH}\); the molar mass of styrene is \(104.14 \mathrm{~g} / \mathrm{mol}\). What number of \(\mathrm{H}\) atoms are present in a \(2.00-\mathrm{g}\) sample of styrene?

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.
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