试题与答案

NaCl和CaCl2的固体混合物14g,加入到96g碳酸钠溶液中恰好完全反应,待

题型:问答题

题目:

NaCl和CaCl2的固体混合物14g,加入到96g碳酸钠溶液中恰好完全反应,待反应完全后,过滤、称得滤液的质量为100g

求:(1)固体混合物中CaCl2的质量是多少?

(2)所得滤液中溶质的质量分数是多少?

答案:

(1))由题意可知,滤液的总质量与生成沉淀的质量之和应等于反应前所有混合物的总质量.所以生成沉淀的质量=14g+96g-100g=10g;

设固体混合物中氯化钙的质量为x,生成NaCl的质量为y

CaCl2+Na2CO3=2NaCl+CaCO3

111          117    100

x             y     10g

100
111
=
10g
x
100
117
=
10g
y

解之得:x=11.1g,y=11.7g.

(2)所得滤液即为NaCl溶液,溶液的总质量为100克,溶质NaCl的质量为14g-11.1g+11.7g=14.6g

所以滤液中溶质的质量分数=

14.6g
100g
×100%=14.6%

答:固体混合物中CaCl2的质量是11.1g,所得滤液中溶质的质量分数是14.6%.

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