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# 200cm$$^3$$ of 0.50mol/dm$$^3$$ solution of calcium hydrogen trioxocarbonate (IV)...

### Question

200cm$$^3$$ of 0.50mol/dm$$^3$$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

A) 10g

B) 25g

C) 20g

D) 15g

### Explanation:

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$$_3$$.
$$\frac{1}{0.10} = \frac{100}{x}$$
$$x = 100 \times 0.10 = 10 g$$

## Dicussion (1)

• 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$$_3$$.
$$\frac{1}{0.10} = \frac{100}{x}$$
$$x = 100 \times 0.10 = 10 g$$