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Chapter 2: Problem 9

In one experiment, 2.18 g sodium was allowed to react with \(16.12 \mathrm{g}\) chlorine. All the sodium was used up, and 5.54 g sodium chloride (salt) was produced. In a second experiment, 2.10 g chlorine was allowed to react with \(10.00 \mathrm{g}\) sodium. All the chlorine was used up, and 3.46 g sodium chloride was produced.Show that these results are consistent with the law of constant composition.

Short Answer

Expert verified
The percentages by mass of sodium and chlorine in sodium chloride for Experiment 1 are approximately 39.35% and 60.65%, respectively. For Experiment 2, these values are approximately 39.31% and 60.69% respectively. This shows that the results are consistent with the law of constant composition.

Step by step solution

01

Calculating the mass of Chlorine in Experiment 1

From Experiment 1, the mass of sodium chloride produced is 5.54 g and the mass of sodium used up is 2.18 g. To get the mass of chlorine, subtract the mass of sodium from the total mass of sodium chloride, i.e., \(5.54 - 2.18 = 3.36\) g.
02

Calculating the percentages of Sodium and Chlorine in Experiment 1

The percentage of sodium (by mass) in the compound is given as \(\frac{2.18}{5.54} \times 100 = 39.35 \% \). The percentage of chlorine (by mass) in the compound is \(\frac{3.36}{5.54} \times 100 = 60.65 \% \).
03

Calculating the mass of Sodium in Experiment 2

From Experiment 2, the mass of sodium chloride produced is 3.46 g and the mass of chlorine used up is 2.10 g. The mass of sodium is found by subtracting the mass of chlorine from the total mass of sodium chloride, i.e., \(3.46 - 2.10 = 1.36\) g.
04

Calculating the percentages of Sodium and Chlorine in Experiment 2

The percentage of sodium (by mass) in the compound is \(\frac{1.36}{3.46} \times 100 = 39.31 \% \). The percentage of chlorine (by mass) in the compound is \(\frac{2.10}{3.46} \times 100 = 60.69 \% \).
05

Comparing the Results

The percentages by mass of sodium and chlorine in sodium chloride are approximately the same in both experiments. This shows that the results are consistent with the law of constant composition.

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