HINT: So if you look at the periodic table and the oxidation state of the carbonate anion...and that coefficient on that product 2XClyou will be able to tell which GROUP your answer lies in....that limits what you are looking for...

The answer of the group of X is given in the text. --> "alkali metal"

Let me try to explain:

n(X

_{2}CO

_{3}):n(CaCl

_{2}):n(CaCO

_{3})=1:1:1

n(CaCl

_{2})=c(CaCl

_{2})*V(CaCl

_{2})=30 ml * 0.5 mmol/ml = 15 mmol

With m(X

_{2}CO

_{3}) and n(CaCl

_{2}), we can calculate the minimum molar mass of the carbonate that's required for a complete precipitation:

M

_{min}(X

_{2}CO

_{3})=m(X

_{2}CO

_{3})/n(X

_{2}CO

_{3})=115.5 mg/15 mmol = 7.7 g/mol

Every possible carbonate has a molar mass greater than 7.7, so we can be sure that X

_{2}CO

_{3} reacts completely to CaCO

_{3}. The following calculation would have worked without this step, too (but we didn't know that before).

n(CaCO

_{3})=m(CaCO

_{3})/M(CaCO

_{3})=50.1 mg / 100.1 g/mol = 0.5 mmol

We know: n(X

_{2}CO

_{3})=n(CaCO

_{3}) and M(X

_{2}CO

_{3})=60+2x so

M(X

_{2}CO

_{3})=m(X

_{2}CO

_{3})/n(X

_{2}CO

_{3}) <==> 60+2x=115.5 / 0.5 mmol

==>

**x=M(X)=85.5 g/mol**