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WAEC Chemistry 2012 Paper

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Question : 15

Total: 44

Consider the following reaction equation: CaCO

_{3(s)}

_{(aq)}

_{2(aq)}

_2

_{(l)}

_{2(g)}

. What mass of CaCI

_2

would be obtained when 25.0g of CaCO

_{3(s)}

is reacted with excess HCI

_{(aq)}

{}_3 = 100; CaCI_2

= 111]

4.00g
4.44 g
18.9 g
27.8 g

Solution:

Concept:

Use the mole ratio from the balanced equation to convert the given mass of

\mathrm{CaCO_3}

into the mass of

\mathrm{CaCl_2}

formed.

Explanation:

The balanced equation is:

\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)}

From the equation,

1

mole of

\mathrm{CaCO_3}

gives

1

mole of

\mathrm{CaCl_2}

.

Molar mass of

\mathrm{CaCO_3} = 100\ \mathrm{g\,mol^{-1}}

and molar mass of

\mathrm{CaCl_2} = 111\ \mathrm{g\,mol^{-1}}

.

So,

100\ \mathrm{g}

of

\mathrm{CaCO_3}

produces

111\ \mathrm{g}

of

\mathrm{CaCl_2}

.

Therefore, mass of

\mathrm{CaCl_2}

from

25.0\ \mathrm{g}

of

\mathrm{CaCO_3}

is:

\frac{25.0 \times 111}{100} = 27.75\ \mathrm{g} \approx 27.8\ \mathrm{g}

Answer:

27.8\ \mathrm{g}

(Option D)

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