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Chapter 2: Problem 6

Within the limits of experimental error, show that the law of conservation of mass was obeyed in the following experiment: \(10.00 \mathrm{g}\) calcium carbonate (found in limestone) was dissolved in 100.0 mL hydrochloric acid \((d=1.148 \mathrm{g} / \mathrm{mL}) .\) The products were \(120.40 \mathrm{g}\) solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L carbon dioxide gas \((d=1.9769 \mathrm{g} / \mathrm{L})\)

Short Answer

Expert verified
The law of conservation of mass is confirmed in this experiment as the total mass of the reactants (124.8 g) is approximately equivalent to the total mass of the products (124.78 g), within the limits of experimental error.

Step by step solution

01

Calculate the mass of the hydrochloric acid

First, recall that density is mass/volume. Therefore, the mass of the hydrochloric acid is equal to its density times its volume, which is \(1.148 \mathrm{g/mL} \times 100.0 \mathrm{mL} = 114.8 \mathrm{g}\).
02

Calculate the total mass of the reactants

Add the mass of the calcium carbonate and the mass of the hydrochloric acid to find the total mass of the reactants. This equals \(10.00 \mathrm{g} + 114.8 \mathrm{g} = 124.8 \mathrm{g}\).
03

Calculate the mass of the carbon dioxide

Again, using the density is mass/volume formula, the mass of the carbon dioxide is \(1.9769 \mathrm{g/L} \times 2.22 \mathrm{L} = 4.38 \mathrm{g}\).
04

Calculate the total mass of the products

Add the mass of the solution and the mass of the carbon dioxide to find the total mass of the products. This equals \(120.40 \mathrm{g} + 4.38 \mathrm{g} = 124.78 \mathrm{g}\).
05

Compare the total mass of the reactants with the total mass of the products

Comparing the total mass of reactants and the total mass of products, they are \(124.8 \mathrm{g}\) and \(124.78 \mathrm{g}\), respectively, which within the limits of experimental error, are equivalent thus confirming the law of conservation of mass.

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