Question

1 gram of a mixture of caco3 and nacl reacts completely with 120 ml of 0.1N HCl. what is the percentage of NaCl?

Answer 1

Answer on Question #43191 - Chemistry - Inorganic chemistry

Question:

1 gram of a mixture of $\mathrm{CaCO_3}$ and NaCl reacts completely with 120ml of 0.1N HCl. What is the percentage of NaCl?

Solution:

HCl reacts only with one component ($\mathrm{CaCO_3}$) of the two components of mixture ($\mathrm{CaCO_3 + NaCl}$) according to the equation:

$$\mathrm{CaCO_3} + 2\mathrm{HCl} = \mathrm{CaCl_2} + \mathrm{CO_2} + \mathrm{H_2O}$$

If we calculate the mass of $\mathrm{CaCO_3}$ that reacts with HCl, we can determine the mass and percentage of NaCl.

Our path:

1. 0,1N HCl means that is 0,1 equivalents of HCl per 1 liter of solution,

0,1 eq HCl/1 l,

$120\mathrm{ml} = 0,12\mathrm{l}$,

$0,12\mathrm{l} \times 0,1\mathrm{eq}$ HCl/1 l = 0,012 eq HCl

2. In case of HCl 1 equivalent = 1 mole,

then 0,012 equivalents = 0,012 moles

3. From the balanced equation we see that 1 mole of $\mathrm{CaCO_3}$ reacts with 2 moles of HCl:

$1\mathrm{mol}\ \mathrm{CaCO_3}/2\mathrm{mol}\ \mathrm{HCl}$;

then 0,012 moles of HCl react with:

$0,012\mathrm{mol}\ \mathrm{HCl} \times 1\mathrm{mol}\ \mathrm{CaCO_3}/2\mathrm{mol}\ \mathrm{HCl} = 0,006\mathrm{mol}\ \mathrm{CaCO_3}$;

4. Mass of 1 mole of $\mathrm{CaCO_3}$ is $40 + 12 + 16 \times 3 = 100\mathrm{g}$

$100\mathrm{g}\ \mathrm{CaCO_3}/1\mathrm{mol}\ \mathrm{CaCO_3}$;

mass of 0,006 moles of $\mathrm{CaCO_3}$ is

$0,006\mathrm{mol}\ \mathrm{CaCO_3} \times 100\mathrm{g}\ \mathrm{CaCO_3}/1\mathrm{mol}\ \mathrm{CaCO_3} = 0,6\mathrm{g}\ \mathrm{CaCO_3}$;

5. Mass of \mathrm{NaCl} = (\text{mass of mixture}) - (\text{mass of\mathrm{CaCO_3}})

$1 - 0,6 = 0,4\mathrm{g}$

percentage of $\mathrm{NaCl} = (\text{mass of NaCl}) / (\text{mass of mixture}) \times 100\%$

$\mathrm{w_{NaCl}} = 0,4 / 1 \times 100\% = 40\%$

Answer:

The percentage of NaCl is $40\%$.

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