Na2CO3+CaCl2═CaCO3↓+2NaCl
106 111 100 117
53g×2% x z y
106 |
53g×2% |
111 |
x |
100 |
z |
117 |
y |
解得:x=1.11g,y=1.17g,z=1g
(2)反应后所得溶液中溶质的质量分数=
25g−1.11g+1.17g |
25g+173g+53g−1g |
答:(1)原食盐中有氯化钙的质量是1.11g;
(2)反应后所得溶液中溶质的质量分数是10%.
106 |
53g×2% |
111 |
x |
100 |
z |
117 |
y |
25g−1.11g+1.17g |
25g+173g+53g−1g |