Equation: Ca(HCO\(_3\))\(_2\) \(\to\) CaCO\(_3\) (precipitate) + H\(_2\)O + CO\(_2\) (in the presence of heat)
V = 200 cm\(^3\) = 0.2 dm\(^3\)
C = 0.5 mol/dm\(^3\) = 0.5M
N = CV \(\implies\) N = 0.5 \(\times\) 0.2
= 0.1 mole
From the equation 1 mole of Ca(HCO\(_3\))\(_2\) gives 100g of CaCO\(_3\)
(1 mole CaCO\(_3\) = 40 + 12 + (3 x 16) = 100g\)
0.1 mole of Ca(HCO\(_3\))\(_2\) gives x g of CaCO\(_3\).
\(\frac{1}{0.10} = \frac{100}{x}\)
\(x = 100 \times 0.10 = 10 g\)