`4*4 g` of an oxide of nitrogen gives `2*24 L` of nitrogen and 60 g of another oxide of nitrogen gives `22*4 L` of nitrogen at STP. The data illustrates

Options

A. Law of conservation of mass
B. Law of constant proportion
C. Law of multiple proportions
D. Law of reciprocal proportions.

Explanation

To solve the problem step-by-step, we will analyze the data provided for the two oxides of nitrogen and determine which law is illustrated by the given information.

### Step 1: Analyze the first oxide of nitrogen
We are given:
- Mass of the first oxide of nitrogen = 4.4 g
- Volume of nitrogen produced = 2.24 L

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Therefore, we can calculate the number of moles of nitrogen produced.

**Calculation:**
\[ \text{Number of moles of } N_2 = \frac{\text{Volume of } N_2}{\text{Molar volume at STP}} = \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mol} \]

### Step 2: Calculate the mass of nitrogen in the first oxide
The molar mass of nitrogen (N) is 14 g/mol, so the mass of nitrogen produced can be calculated as follows:

**Calculation:**
\[ \text{Mass of } N = \text{Number of moles} \times \text{Molar mass} = 0.1 \, \text{mol} \times 28 \, \text{g/mol} = 2.8 \, \text{g} \]

### Step 3: Analyze the second oxide of nitrogen
We are given:
- Mass of the second oxide of nitrogen = 60 g
- Volume of nitrogen produced = 22.4 L

**Calculation:**
\[ \text{Number of moles of } N_2 = \frac{22.4 \, \text{L}}{22.4 \, \text{L/mol}} = 1 \, \text{mol} \]

### Step 4: Calculate the mass of nitrogen in the second oxide
Using the molar mass of nitrogen again, we can find the mass of nitrogen produced:

**Calculation:**
\[ \text{Mass of } N = 1 \, \text{mol} \times 28 \, \text{g/mol} = 28 \, \text{g} \]

### Step 5: Calculate the mass of oxygen in both oxides
**For the first oxide:**
- Total mass of the first oxide = 4.4 g
- Mass of nitrogen = 2.8 g
- Mass of oxygen = Total mass - Mass of nitrogen = 4.4 g - 2.8 g = 1.6 g

**For the second oxide:**
- Total mass of the second oxide = 60 g
- Mass of nitrogen = 28 g
- Mass of oxygen = Total mass - Mass of nitrogen = 60 g - 28 g = 32 g

### Step 6: Determine the ratio of the masses of oxygen
Now we can compare the masses of oxygen in both oxides:

- Mass of oxygen in the first oxide = 1.6 g
- Mass of oxygen in the second oxide = 32 g

To find the ratio of the masses of oxygen, we can set up the following:

\[ \text{Ratio} = \frac{1.6 \, \text{g}}{32 \, \text{g}} = \frac{1}{20} \]

### Step 7: Conclusion
According to the law of multiple proportions, when two elements form different compounds, the ratios of the masses of one element that combine with a fixed mass of the other element can be expressed in small whole numbers. Here, we see that the ratio of the masses of oxygen in the two oxides is 1:20, which illustrates the law of multiple proportions.

### Final Answer
The data illustrates the **Law of Multiple Proportions**.

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