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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

Options

A) 10g

B) 25g

C) 20g

D) 15g

The correct answer is A.

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

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

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