1.00 g of oxygen combine with 0.126 g of hydrogen to form `H_2O` .1.00 g of nitrogen combine with 0.216g of hydrogen to form `NH_3`. Predict the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide.

Answer

To solve the problem, we will use the law of reciprocal proportions. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- 1.00 g of oxygen combines with 0.126 g of hydrogen to form water (H₂O).
- 1.00 g of nitrogen combines with 0.216 g of hydrogen to form ammonia (NH₃).

2. **Use the Law of Reciprocal Proportions**:
- According to the law, if a certain amount of one element (hydrogen) reacts with two different elements (oxygen and nitrogen), we can find the ratio of the elements involved.

3. **Calculate the Amount of Oxygen Required for 0.216 g of Hydrogen**:
- We know from the first reaction that 0.126 g of hydrogen reacts with 1.00 g of oxygen.
- We need to find out how much oxygen reacts with 0.216 g of hydrogen.
- Set up the proportion:
\[ \text{If } 0.126 \text{ g H} \text{ reacts with } 1.00 \text{ g O, then } 0.216 \text{ g H reacts with } x \text{ g O.} \]
- This can be expressed as:
\[ x = \frac{0.216 \text{ g H} \times 1.00 \text{ g O}}{0.126 \text{ g H}}. \]

4. **Calculate x**:
- Performing the calculation:
\[ x = \frac{0.216 \times 1.00}{0.126} = 1.7143 \text{ g O} \approx 1.71 \text{ g O}. \]

5. **Determine the Ratio of Nitrogen to Oxygen**:
- From the second reaction, we know that 0.216 g of hydrogen reacts with 1.00 g of nitrogen.
- Now we have determined that 0.216 g of hydrogen reacts with approximately 1.71 g of oxygen.
- Therefore, the mass ratio of nitrogen to oxygen can be established as:
\[ \text{Mass ratio of N to O} = \frac{1.00 \text{ g N}}{1.71 \text{ g O}}. \]

6. **Conclusion**:
- From the above calculations, we conclude that 1.71 g of oxygen will combine with 1.00 g of nitrogen to form an oxide.

### Final Answer:
Hence, the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide is **1.71 g**.