有一种石灰石样品的成分是CaCO3和SiO2.课外小组同学将100g盐酸分5次加入到35g石灰石样品中(已知SiO2不

2025-03-17 20:56:38
推荐回答(1个)
回答1:

(1)由第1次实验可得知,20g稀盐酸能反应掉5g碳酸钙;因此,第2次加入20g稀盐酸时,固体质量会继续减少5g,所剩余固体的质量=30g-5g=25g;
故答案为:25;
(2)石灰石样品中碳酸钙的质量分数=

35g?15g
35g
×100%≈57.1%
答:石灰石样品中碳酸钙的质量分数为57.1%;
(3)100g稀盐酸完全反应消耗碳酸钙的质量=5g×
100g
20g
=25g
设100g稀盐酸与25g碳酸钙恰好完全反应后生成氯化钙的质量为x,放出二氧化碳的质量为y
CaCO3+2HCl═CaCl2+H2O+CO2
100         111       44
25g          x         y
100
25g
111
x
  x=27.75g
100
25g
44
y
  y=11g
27.75g氯化钙可形成10%的溶液的质量=
27.75g
10%
=277.5g
则需要加入水的质量=277.5-(100g+25g-11g)=163.5g
答:还需要向滤液中加入水163.5g;
(4)图象的起点为35g;图象的折点:当第4次加入20g稀盐酸即加入80g稀盐酸时,碳酸钙完全反应,剩余固体的质量为15g;图象的趋势:再加入稀盐酸,固体质量不变;
故答案为:

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