分析某硅酸盐得如下结果:SiO2 45.7%Al2O3 &nb
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:化学作业 时间:2024/08/11 20:24:49
分析某硅酸盐得如下结果:SiO2 45.7%Al2O3 38.41%K2O 12.1%H2O 4.42% ,则该盐的化学式为 ( )A.K2O · Al2O3 · 2SiO2 · 2H2OB.K2O · Al2O3 · 4SiO2 · 2H2OC.K2O · 2Al2O3 · 5SiO2 · 2H2OD.K2O · 3Al2O3 · 6SiO2 · 2H2O 要过程!
![分析某硅酸盐得如下结果:SiO2 45.7%Al2O3 &nb](/uploads/image/z/13906113-33-3.jpg?t=%E5%88%86%E6%9E%90%E6%9F%90%E7%A1%85%E9%85%B8%E7%9B%90%E5%BE%97%E5%A6%82%E4%B8%8B%E7%BB%93%E6%9E%9C%3ASiO2%26nbsp%3B%26nbsp%3B+%26nbsp%3B45.7%25Al2O3%26nbsp%3B%26nbsp%3B%26nb)
首先SiO2的摩尔质量是60g/mol,Al2O3的是102g/mol,K20 94g/molH2O18g/mol,
然后SiO2:Al2O3:K2O:H2O的物质的量是45.7%/60 :38.41%/102 :12.1%/94 :4.42%/18 =0.76:0.38:0.13:0.25,所以他们之间的比是6:3:1:2,所以选D
然后SiO2:Al2O3:K2O:H2O的物质的量是45.7%/60 :38.41%/102 :12.1%/94 :4.42%/18 =0.76:0.38:0.13:0.25,所以他们之间的比是6:3:1:2,所以选D