有一包白色固体粉末,可能含有CuSO4、BaCl2、Na2SO4、Na2CO3、NaCl中的一种或几种,某同学对其组成进行了

2025-04-04 15:49:35
推荐回答(1个)
回答1:

白色粉末加水得到了无色溶液,故一定不含有硫酸铜;白色沉淀A加入硝酸能产生无色气体C,C能使澄清的石灰水变浑浊,则C是二氧化碳,D是碳酸钙,说明A中含有碳酸根离子,而能生成碳酸盐沉淀的物质是氯化钡和碳酸钠,故白色固体中一定含有氯化钡和碳酸钠;A中加入足量的稀硝酸,仍然有白色沉淀B,故B是不溶于稀硝酸的沉淀硫酸钡,故A中还含有硫酸钠;无色溶液加入硝酸银溶液能产生白色沉淀,其中的氯离子可能是来自含有的氯化钠也可能是来自氯化钡反应后生成的氯化钠,无法确定氯化钠的存在;
(1)这包白色粉末中一定含有氯化钡、硫酸钠、碳酸钠,一定不含有硫酸铜,可能含有氯化钠,故填:BaCl2、Na2SO4、Na2CO3;CuSO4,NaCl;
(2)白色沉淀B是硫酸钡,溶液F中含有硝酸钠,故填:BaSO4,NaNO3
(3)碳酸钡能与硝酸反应生成硝酸钡、水和二氧化碳,故填:BaCO3+2HNO3═Ba(NO32+H2O+CO2↑.

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