Question

in an exprinment 5g CaCO3 on heating gave 2.8g CaO and 1120ml 0f CO2 at STP show that these result are in agreement with law of conservation of mass

Answer 1

Answer on Question #43453 - Chemistry - Other

Question:

in an experiment 5g CaCO3 on heating gave 2.8g CaO and 1120ml Of CO2 at STP show that these result are in agreement with law of conservation of mass

Answer:

Balanced equation for $\mathrm{CaCO_3}$ heating:

$$\mathrm{CaCO_3} = \mathrm{CaO} + \mathrm{CO_2} \uparrow$$

The law of conservation of mass states that the mass of the system must remain constant over time, as system mass cannot change quantity if it is not added or removed. The law implies that during any chemical reaction in an isolated system, the total mass of the reactants or starting materials must be equal to the mass of the products.

To check if the result of the experiment is in agreement with this law, we should calculate the mass of $\mathrm{CO_2}$ produced.

Number of moles of $\mathrm{CO_2}$ equals:

$$n(\mathrm{CO_2}) = \frac{V}{V_m} = \frac{1120\,\mathrm{ml}}{22.4\,\mathrm{L/mol}} = \frac{1.120\,\mathrm{ml}}{22.4\,\mathrm{L/mol}} = 0.05\,\mathrm{mol}$$

Then the mass of $\mathrm{CO_2}$ produced equals:

$$m(\mathrm{CO_2}) = n(\mathrm{CO_2}) \cdot M(\mathrm{CO_2}) = 0.05\,\mathrm{mol} \cdot 44\,\frac{\mathrm{g}}{\mathrm{mol}} = 2.2\,\mathrm{g}$$

The total mass of the products is:

$$m = m(\mathrm{CaO}) + m(\mathrm{CO_2}) = 2.8 + 2.2 = 5.0\,\mathrm{g}$$

We see that heating of $5\,\mathrm{g}$ of $\mathrm{CaCO_3}$ produces $5\,\mathrm{g}$ of the products ($2.8\,\mathrm{g}$ of $\mathrm{CaO}$ and $2.2\,\mathrm{g}$ of $\mathrm{CO_2}$), i.e. the total mass of the reactants is equal to the mass of the products. So, the result of the experiment is in agreement with law of conservation of mass.

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