问题 填空题

(3分)实验室有一瓶失去标签的黑色粉末A和~瓶标签不全的无色溶液B(如图l8)。小明想知道可能是什么物质,他各取少量于试管中混合,发现产生气泡;又查阅资料得知初中化学常见黑色粉末有:Cu0、  Mn02、Fe304、铁粉和炭粉。

(1)你认为黑色粉末A是  ▲ ;无色溶液B是  ▲ 

(2)请写出相应的化学反应方程式  ▲ 

答案

答案l(1)Fe(或铁粉)(1分)H2SO4(或稀硫酸)(1分)

(2)Fe+ H2SO4===FeSO4 +H2↑ ………………(1分)

(说明:化学式书写错误不给分;没有气体符号、未用等号或同时出现以上情况的均扣0.5分)

答案2:(1)MnO2(或二氧化锰)(1分)  H2O2(或双氧水)(1分)

(2) 2H2O22H2O+O2↑   ……………….(1分)

(说明:化学式书写错误不给分;未标催化剂、未配平、没有气体符号、未用等号或同时出现以上情况的均扣0.5分)

分析:(1)根据该粉末为黑色且能与某种溶液反应放出气体的变化特点,利用所提示的黑色粉末的性质,对黑色粉末的可能性进行分析;

(2)根据所确定的黑色粉末与无色溶液反应,写出该反应的化学方程式.

解:(1)在所列出的黑色粉末中,炭粉不能与溶液发生反应,而CuO、Fe304与溶液反应时不会产生气体,只有Mn02能与过氧化氢溶液混合放出氧气、铁粉能与稀硫酸或稀盐酸反应放出氢气;因此黑色粉末A可能为Mn02或铁粉,则无色溶液B对应地为过氧化氢溶液或稀硫酸(稀盐酸);

(2)黑色粉末铁粉与稀硫酸反应生成硫酸亚铁和氢气,反应的化学方程式为Fe+H2SO4═FeSO4+H2↑;

黑色粉末Mn02与过氧化氢溶液混合反应的化学方程式为2H2O22H2O+O2

答案1:(1)Fe;H2SO4

(2)Fe+H2SO4═FeSO4+H2↑.

答案2:(1)MnO2; H2O2

(2)2H2O22H2O+O2

点评:化学方程式书写时,正确书写反应中各物质的化学式、配平并正确标明反应的条件及气体(或沉淀)符号,这些都将影响化学方程式的正确性.

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