不少家庭养观赏鱼，有时会在鱼缸里放点“鱼浮灵”，可使水中的含氧量迅速增加．请你和小欣同学一起完成这
推荐回答（1个）
探究一：【设计方案】①过碳酸氢钠溶于水，在二氧化锰的催化作用下，迅速的分解产生氧气，能使带火星的木条复燃；
②常用稀盐酸来检验碳酸根离子的存在，向待测物质中加入稀盐酸，若能产生使澄清的石灰水变浑浊的气体，则证明该物质中含有碳酸根离子；
【结果分析】由质量守恒定律知，化学式X中含有两个钠原子、一个碳原子和四个氧原子，所以X化学式是Na2CO4．
【实验反思】（1）仪器a、b的名称为分液漏斗；
（2）过氧化氢在二氧化锰的催化作用下分解为水和氧气；
（3）要将产生的气体再通入澄清的石灰水，观察是否变浑浊才能进一步证明是否是二氧化碳气体；
（4）因为其遇水会发生反应产生氧气，故应密封保存，置于阴凉处；
探究二：
方案一：产生的氧气会从装置中溢出，故装置和试剂减少的质量就是生成氧气的质量；产生的氧气中会混有水蒸气，故要用碱石灰来吸收其中的水蒸气；
取一份过碳酸钠晶体溶于水，和二氧化锰固体按照如图2（方法二）装置进行装配，要使测定结果准确，必须调整洗气瓶与量筒液面等高，氧气的质量等于其体积与密度的乘积；
方案二：
（1）过量的氯化钙溶液才能保证将碳酸钠转化为沉淀；
（2）过滤得到的沉淀表面会有氯化钙、氯化钠等杂质，所以要洗涤干净；
（3）如果沉淀没有洗涤干净，则其表面一定有氯离子，故可用硝酸银溶液来进行检验；
【实验结果】设产生5g碳酸钙沉淀，需要碳酸钠的质量为x
Na2CO3+CaCl2=CaCO3↓2NaCl
106 100
x 5g
＝
x=5.3g
则7.85克过碳酸钠晶体，含有H2O2的质量为：7.85g-5.3g=2.55g
则：＝，＝
过碳酸钠晶体（xNa2CO3?yH202）的化学式为：2Na2CO3?3H202．
【实验反思】根据以上计算过程分析可知，若CaCO3沉淀过滤后不洗涤，则x：y的值将偏大．
故答案为：
探究一：【设计方案】
| 实验 | 主要操作步骤 | 主要实验现象 |
| ① | MnO2
| 有气泡逸出，能使带火星木条复燃 |
| ② | 稀盐酸 | |
【结果分析】Na
2CO
4．
【实验反思】（1）分液漏斗；
（2）2H
2O
2 2H
2O+O
2↑
（3）生成的气体通入澄清的石灰水；
（4）密封，置阴凉处．
探究二：【设计实验】方案一：
【方法一】m
1-m
2；吸收氧气中的水蒸气，防止水蒸气逸出装置．
【方法二】洗气瓶与量筒液面等高；该温度下氧气的密度．
方案二：（1）将 Na
2CO
3 完全转化为沉淀；
（2）除去表面可溶性杂质 CaCl
2、NaCl；
（3）取最后一滴洗涤液，滴加AgNO
3若无沉淀出现，则表示已经洗净．
【实验结果】
2Na
2CO
3?3H
20
2【实验反思】偏大．
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