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某铁的氧化物，用

7 \mathrm{~mol} / \mathrm{L}

的盐酸

100 \mathrm{~mL}

在一定条件下恰好完全溶解，所得溶液...

Mar 19, 2024

某铁的氧化物，用

7 \mathrm{~mol} / \mathrm{L}

的盐酸

100 \mathrm{~mL}

在一定条件下恰好完全溶解，所得溶液再通入

0.56 \mathrm{~L}

标准状况下的氯气时，刚好使溶液中

\mathrm{Fe}^{2+}

完全转化为

\mathrm{Fe}^{3+}

。则该氧化物的化学式可表示为( ) A.

\mathrm{FeO}

\mathrm{Fe}_{3} \mathrm{O}_{4}

\mathrm{Fe}_{4} \mathrm{O}_{5}

\mathrm{Fe}_{5} \mathrm{O}_{7}

Key Concept

Stoichiometry and redox reactions

Explanation

To determine the formula of the iron oxide, we need to consider the stoichiometry of the reaction with hydrochloric acid and the subsequent redox reaction with chlorine gas. The amount of

\mathrm{Fe}^{2+}

produced from the oxide by hydrochloric acid must be equal to the amount of

\mathrm{Fe}^{2+}

oxidized to

\mathrm{Fe}^{3+}

by chlorine gas. From the given data,

100 \mathrm{~mL}

7 \mathrm{~mol} / \mathrm{L}

HCl is used, which is

0.7 \mathrm{~mol}

of HCl. Chlorine gas at standard conditions has a molar volume of

22.4 \mathrm{~L/mol}

, so

0.56 \mathrm{~L}

\mathrm{Cl}_2

0.025 \mathrm{~mol}

\mathrm{Cl}_2

. The reaction of

\mathrm{Fe}^{2+}

with

\mathrm{Cl}_2

is in a 2:1 molar ratio, so

0.025 \mathrm{~mol}

\mathrm{Cl}_2

would react with

0.05 \mathrm{~mol}

\mathrm{Fe}^{2+}

. Since the

\mathrm{Fe}^{2+}

comes from the iron oxide reacting with HCl, and the ratio of HCl to

\mathrm{Fe}^{2+}

in the reaction with

\mathrm{FeO}

\mathrm{Fe}_3\mathrm{O}_4

is 2:1 or 6:1 respectively, we can calculate the moles of iron in the oxide. For

\mathrm{Fe}_3\mathrm{O}_4

0.7 \mathrm{~mol}

of HCl would produce

0.7 \mathrm{~mol} \div 6 = 0.1167 \mathrm{~mol}

\mathrm{Fe}^{2+}

, which is more than enough to react with the available chlorine. However, for

\mathrm{FeO}

0.7 \mathrm{~mol}

of HCl would produce

0.35 \mathrm{~mol}

\mathrm{Fe}^{2+}

, which is too much for the available chlorine. Therefore, the correct formula is

\mathrm{Fe}_3\mathrm{O}_4