Are you sure that was 0.7374 g Ca? or was it CaCO3. The answer below is for 0.7374 g Ca.
0.7374 x (molar mass Ca citrate/atomic mass Ca) = ?
For other metals look for other carbonates that are insoluble; ;i.e., Mg might be one.
What mass of calcium citrate, Ca3(C6 H5 O7)2, would a tablet manufacturer need to use to provide the same amount of Ca that you found in your tablet? (show calculations)
The mass of Calcium for the experiment was .7374g
We assumed that any solid that formed when Na2CO3 was added was CaCO3 . This is not necessarily true. There may have been other cation in the tablet that could also have reacted with Na2CO3 to form a precipitate. Give any two other cations that might have have been present in the tablet that would have formed a precipitate when you added Na2CO3 solution. There are many correct answers to this question.
1 answer
2 answers