摘要:25. (1)根据题意 n(HCl) = 12 mol/L×0.2 L = 2.4 mol n(MnO2) = 17.4 g÷87 g/mol = 0.2 mol MnO2 + 4 HCl(浓) △ MnCl2 + Cl2↑+ 2H2O n(MnO2)∶n(HCl) = 2.4 mol∶0.2 mol = 12∶1 > 4∶1 所以浓HCl过量.应根据MnO2计算. 根据制备氯气的化学方程式 n(Cl\2) = n(MnO2) = 0.2 mol 又因为:2Cl2 + 2Ca(OH)2 = Ca(ClO)2 + CaCl2 + 2H2O n[Ca(ClO)2] = 1/2 n(Cl2) = 1/2×0.2 mol = 0.1 mol m[Ca(ClO)2] = 143 g/mol × 0.1 mol = 14.3 g (2)①随着反应进行.温度升高.会产生副产物Ca(ClO3)2, 6Cl2 + 6Ca(OH)2 △ 5CaCl2 + Ca(ClO3)2 + 6H2O ②Cl2未与石灰乳完全反应.残余Cl2被NaOH溶液吸收 Cl2 + 2NaOH = NaCl + NaClO + H2O

网址:http://m.1010jiajiao.com/timu_id_3694830[举报]

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网