0.0859g of compound X which contains a group II metal dissolves in water to produce colourless solution.
excess acidifed silver nitrate added. 0.308g of white precipitate (Y) was formed,
What is Y? My answer = AgCl
Compound X? My answer = CaCl2
Is it Ca?????????? Watch
- Thread Starter
- 14-04-2013 20:33
- 14-04-2013 21:21
I think it would have to be a lighter group 2 metal than Ca.
0.308g of AgCl would be 0.00215 mol. Halving that because a grp2 compound would have two Cl, you need 0.00108 ish moles of your group 2 compound to produce this. Try the calculations with Mg and Be and see which fits.