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已知: 将

\mathrm{Cl}_{2}

通入适量

\mathrm{KOH}

溶液，产物中可能有 $\mathrm{KCl} 、 \mathrm{KC...

Mar 19, 2024

已知: 将

\mathrm{Cl}_{2}

通入适量

\mathrm{KOH}

溶液，产物中可能有

\mathrm{KCl} 、 \mathrm{KClO} 、 \mathrm{KClO}_{3}

，且

\frac{c\left(\mathrm{Cl}^{-}\right)}{c\left(\mathrm{ClO}^{-}\right)}

的值与温度高低有关。当

n(\mathrm{KOH})=a \mathrm{~mol}

时下列有关说法错误的是( ) A. 若某温度下, 反应后

\frac{c\left(\mathrm{Cl}^{-}\right)}{c\left(\mathrm{ClO}^{-}\right)}=11

, 则溶液中

\frac{c\left(\mathrm{ClO}^{-}\right)}{c\left(\mathrm{ClO}_{3}^{-}\right)}=\frac{1}{2}

B. 参加反应的氯气的物质的量等于

\frac{1}{2} a \mathrm{~mol}

C. 改变温度，反应中转移电子的物质的量

n\left(\mathrm{e}^{-}\right)

的范围:

\frac{1}{2} a \mathrm{~mol} \leqslant n\left(\mathrm{e}^{-}\right) \leqslant \frac{5}{6} a \mathrm{~mol}

D. 改变温度，产物中

\mathrm{KClO}_{3}

的最大理论产量为

\frac{1}{7} a \mathrm{~mol}

Solution

Analyze the given ratio: The ratio

\frac{c(\mathrm{Cl}^{-})}{c(\mathrm{ClO}^{-})} = 11

implies that for every 1 mole of

\mathrm{ClO}^{-}

, there are 11 moles of

\mathrm{Cl}^{-}

Determine the incorrect statement: To find the incorrect statement, we need to consider the stoichiometry of the reactions involving

\mathrm{Cl}_{2}

and

\mathrm{KOH}

. The reactions are as follows:

\mathrm{Cl}_{2} + 2\mathrm{KOH} \rightarrow \mathrm{KCl} + \mathrm{KClO} + \mathrm{H}_{2}\mathrm{O}

(at lower temperatures)

3\mathrm{Cl}_{2} + 6\mathrm{KOH} \rightarrow 5\mathrm{KCl} + \mathrm{KClO}_{3} + 3\mathrm{H}_{2}\mathrm{O}

(at higher temperatures) We need to check each statement against these reactions

Check statement A: If

\frac{c(\mathrm{Cl}^{-})}{c(\mathrm{ClO}^{-})} = 11

, then for every 12 moles of

\mathrm{Cl}^{-}

and

\mathrm{ClO}^{-}

combined, there is 1 mole of

\mathrm{ClO}^{-}

. Since the ratio of

\mathrm{ClO}^{-}

\mathrm{ClO}_{3}^{-}

is not given by the reaction stoichiometry directly, we cannot confirm the ratio

\frac{c(\mathrm{ClO}^{-})}{c(\mathrm{ClO}_{3}^{-})} = \frac{1}{2}

without additional information

Check statement B: The reaction at lower temperatures produces

\mathrm{KCl}

and

\mathrm{KClO}

in a 1:1 molar ratio. If

n(\mathrm{KOH}) = a

mol, then the maximum amount of

\mathrm{Cl}_{2}

that can react is

\frac{1}{2}a

mol, because each mole of

\mathrm{Cl}_{2}

reacts with 2 moles of

\mathrm{KOH}

. This statement seems correct based on the stoichiometry

Check statement C: The number of electrons transferred depends on the specific reaction taking place. In the lower temperature reaction, each

\mathrm{Cl}_{2}

molecule gains 2 electrons, while in the higher temperature reaction, each

\mathrm{Cl}_{2}

gains 6 electrons. Therefore, the range of

n(\mathrm{e}^{-})

will depend on the proportion of each product formed and cannot be determined without additional information

Check statement D: The maximum theoretical yield of

\mathrm{KClO}_{3}

would be when all

\mathrm{Cl}_{2}

is converted to

\mathrm{KClO}_{3}

. From the higher temperature reaction, 3 moles of

\mathrm{Cl}_{2}

produce 1 mole of

\mathrm{KClO}_{3}

. Therefore, the maximum amount of

\mathrm{KClO}_{3}

produced from

\frac{1}{2}a

mol of

\mathrm{Cl}_{2}

would be

\frac{1}{6}a

mol, not

\frac{1}{7}a

mol. This statement is incorrect

Answer

Key Concept

Stoichiometry of reactions involving

\mathrm{Cl}_{2}

and

\mathrm{KOH}

Explanation

The incorrect statement is identified by comparing the given ratios and statements with the stoichiometry of the reactions between

\mathrm{Cl}_{2}

and

\mathrm{KOH}

. Statement D is incorrect because it does not match the stoichiometric ratio for the production of

\mathrm{KClO}_{3}