The products obtained during treatment of hard water using Clark's method are

Clark's method is used for removing temporary hardness from water. Temporary hardness is caused by the presence of dissolved bicarbonates of calcium and magnesium, i.e., $$Ca(HCO_3)_2$$ and $$Mg(HCO_3)_2$$.

In Clark's method, a calculated amount of slaked lime, $$Ca(OH)_2$$, is added to the hard water. The reactions that occur are:

For calcium bicarbonate: $$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$$. Here, calcium carbonate ($$CaCO_3$$) precipitates out since it is insoluble in water.

For magnesium bicarbonate: $$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2H_2O$$. In this case, magnesium hydroxide ($$Mg(OH)_2$$) precipitates out because $$MgCO_3$$ is slightly soluble in water. The excess $$Ca(OH)_2$$ converts the magnesium compound to the insoluble $$Mg(OH)_2$$.

So the products obtained during the treatment of hard water using Clark's method are $$CaCO_3$$ and $$Mg(OH)_2$$.

Hence, the correct answer is Option C.