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If [tex]$H_2S$[/tex] contains 94.11% of [tex]$S$[/tex], [tex][tex]$SO_2$[/tex][/tex] contains 50% of [tex]$S$[/tex], and [tex]$H_2O$[/tex] contains 11.11% of [tex][tex]$H$[/tex][/tex], then show that these data illustrate the law of reciprocal proportions.


Sagot :

Certainly! To show that the given data illustrate the law of reciprocal proportions, we need to compare the ratios of the elements involved in different compounds. The law of reciprocal proportions states that the ratio in which one element combines with a fixed mass of another element is the same when they combine with a third element.

Let's start with the given information:

1. [tex]\( H_2S \)[/tex] contains [tex]\( 94.11\% \)[/tex] sulfur (S).
2. [tex]\( SO_2 \)[/tex] contains [tex]\( 50\% \)[/tex] sulfur (S).
3. [tex]\( H_2O \)[/tex] contains [tex]\( 11.11\% \)[/tex] hydrogen (H).

First, let's convert these percentages into fractional values since calculating proportions is more straightforward in fractional form.

### Step 1: Convert Percentages to Fractions
1. Fraction of sulfur in [tex]\( H_2S \)[/tex]:
[tex]\[ \text{Fraction of S in } H_2S = \frac{94.11}{100} = 0.9411 \][/tex]
2. Fraction of sulfur in [tex]\( SO_2 \)[/tex]:
[tex]\[ \text{Fraction of S in } SO_2 = \frac{50}{100} = 0.5 \][/tex]
3. Fraction of hydrogen in [tex]\( H_2O \)[/tex]:
[tex]\[ \text{Fraction of H in } H_2O = \frac{11.11}{100} = 0.1111 \][/tex]

### Step 2: Calculate Fraction of Hydrogen in [tex]\( H_2S \)[/tex]
Since [tex]\( H_2O \)[/tex] contains 11.11% hydrogen, and [tex]\( H_2S \)[/tex] contains two hydrogen atoms:
[tex]\[ \text{Fraction of H in } H_2S = 2 \times 0.1111 = 0.2222 \][/tex]

### Step 3: Calculate Reciprocal Proportions
To apply the law of reciprocal proportions, we need to find the reciprocals of the fractional values:
1. Reciprocal proportion of sulfur in [tex]\( H_2S \)[/tex]:
[tex]\[ \frac{1}{0.9411} = 1.06258633513973 \][/tex]
2. Reciprocal proportion of sulfur in [tex]\( SO_2 \)[/tex]:
[tex]\[ \frac{1}{0.5} = 2.0 \][/tex]
3. Reciprocal proportion of hydrogen in [tex]\( H_2S \)[/tex]:
[tex]\[ \frac{1}{0.2222} = 4.5004500450045 \][/tex]

### Step 4: Comparison and Conclusion
Now we have:
- Reciprocal proportion of sulfur in [tex]\( H_2S \)[/tex]: [tex]\( 1.06258633513973 \)[/tex]
- Reciprocal proportion of sulfur in [tex]\( SO_2 \)[/tex]: [tex]\( 2.0 \)[/tex]
- Reciprocal proportion of hydrogen in [tex]\( H_2S \)[/tex]: [tex]\( 4.5004500450045 \)[/tex]

The ratios thus derived are as follows:
- [tex]\( H_2S \)[/tex] to [tex]\( SO_2 \)[/tex] for sulfur: [tex]\( 1.062586 \)[/tex] vs [tex]\( 2.0 \)[/tex]
- Hydrogen in [tex]\( H_2S \)[/tex]: [tex]\( 4.5004500450045 \)[/tex]

These results show that the reciprocal proportions are consistent with the mass ratios in which sulfur and hydrogen combine to form the respective compounds [tex]\( H_2S \)[/tex] and [tex]\( SO_2 \)[/tex]. They illustrate how these masses are in fixed ratios, providing evidence for the law of reciprocal proportions.

Accordingly, the given data, through the reciprocal proportions calculated, illustrate the law of reciprocal proportions effectively.