工业纯碱主要成分是无水碳酸钠,其中还含有少量氯化钠,为测定工业纯碱中碳酸钠的质量分数,某学生设计了

2025-03-16 19:02:25
推荐回答(1个)
回答1:

(1)不能的,因为这个实验就是通过反应生成二氧化碳,测量二氧化碳的质量来计算碳酸钠质量的,B装置是吸收空气中的二氧化碳的,如果没有B装置,那测量的二氧化碳的质量就会变大,使实验出现误差,同理C装置是吸收气体中水份做干燥剂的,如果没有C装置,碱石灰就会吸收水份,从而使质量比二氧化碳实验质量大,结果偏大的,
生成二氧化碳质量为(m2-m1)
Na2CO3 + 2HCl === 2NaCl + H2O + CO2↑
106 44
X m2-m1
X=(m2-m1)*106/44
质量分数为X/m*100%=53(m2-m1)/22m*100%

亲,有其他题目请另外发问,此问题有疑问,请追问,,以上都是本人自己纯手工做的,有错误,请指出。我是诚心的想帮你,若满意请请点击在下答案旁的"好评",,互相探讨,答题不易,互相理解,请不要随意给差评,谢谢

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