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【题目】H浓度相同的等体积的两份溶液A和B,A为盐酸,B为醋酸,分别和锌反应,若最后仅有一份溶液中存在锌,且放出氢气的体积相等,则下列说法正确的是

反应所需要的时间B>A 开始反应时的速率A>B

参加反应的锌的物质的量A=B 反应过程的平均速率B>A

盐酸中有锌剩余 醋酸中有锌剩余

A.③④⑤ B.③④⑥ C.②③⑤ D.②③⑤⑥

【答案】A

【解析】认识到c(H)相同的两溶液中c(CH3COOH)>c(HCl)是解答本题的关键。锌与盐酸和醋酸的反应本质相同:Zn+2H===Zn2++H2↑。起始时c(H)相同,因此开始时反应速率相同;因放出H2的体积相同,所以参加反应的锌相等;随着反应的进行,醋酸的电离平衡被破坏,向着电离方向移动,故H的变化小于盐酸,所以在反应过程中醋酸中生成氢气的速率大于盐酸中的;锌不可能在醋酸中剩余,因为H浓度相等的同体积醋酸和盐酸相比,醋酸的浓度大于盐酸的浓度。

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