1.00g of oxygen combine with 0.126 g of hydrogen to form H2O. 1.00 g of nitrogen combie with 0.216 g of hydrogen to form ammonia. Predict the weight of oxygen required to combine with 1.00 g nitrogen to form an oxide.i m a student
To predict the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide, we can use the ratios of the elements in the given compounds.
First, let's calculate the molar mass of the elements:
- Oxygen (O) has a molar mass of approximately 16 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.
- Nitrogen (N) has a molar mass of approximately 14 g/mol.
Now, let's analyze the ratios in the given compounds:
1. In water (H2O), 1.00 g of oxygen combines with 0.126 g of hydrogen.
2. In ammonia (NH3), 1.00 g of nitrogen combines with 0.216 g of hydrogen.
Using these ratios, we can determine the number of moles of each element in the compounds:
1. For water, using the molar masses:
- Moles of oxygen = 1.00 g / 16 g/mol = 0.0625 mol (rounded to four decimal places)
- Moles of hydrogen = 0.126 g / 1 g/mol = 0.126 mol
2. For ammonia:
- Moles of nitrogen = 1.00 g / 14 g/mol = 0.0714 mol (rounded to four decimal places)
- Moles of hydrogen = 0.216 g / 1 g/mol = 0.216 mol
Now, we need to find the mole ratio of oxygen to nitrogen in order to predict the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide.
In water (H2O), there is a 1:1 ratio of oxygen to hydrogen, so the mole ratio of oxygen to nitrogen is the same as the mole ratio of hydrogen to nitrogen.
Therefore, the mole ratio of oxygen to nitrogen is:
- Moles of oxygen / Moles of nitrogen = 0.126 mol / 0.0714 mol ≈ 1.7647 (rounded to four decimal places)
Next, let's calculate the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide.
First, determine the number of moles of nitrogen in 1.00 g of nitrogen:
- Moles of nitrogen = 1.00 g / 14 g/mol = 0.0714 mol (rounded to four decimal places)
Now, using the mole ratio calculated earlier, we can determine the number of moles of oxygen required to combine:
- Moles of oxygen = 0.0714 mol * 1.7647 ≈ 0.1264 mol (rounded to four decimal places)
Finally, calculate the weight of oxygen by multiplying the number of moles by the molar mass of oxygen:
- Weight of oxygen = 0.1264 mol * 16 g/mol = 2.0224 g (rounded to four decimal places)
Therefore, approximately 2.0224 grams of oxygen are required to combine with 1.00 g of nitrogen to form an oxide.
To predict the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide, we can use the concept of stoichiometry and the law of conservation of mass.
1. First, we need to determine the molar ratios between the elements involved in each reaction:
In the formation of water (H2O):
- Oxygen (O) combines with Hydrogen (H) in a 1:2 ratio. This means that for every 1.00 g of oxygen, we need 2.00 g of hydrogen.
In the formation of ammonia (NH3):
- Nitrogen (N) combines with Hydrogen (H) in a 1:3 ratio. This means that for every 1.00 g of nitrogen, we need 3.00 g of hydrogen.
2. Now, we need to find the molar mass of each element involved:
- Oxygen (O): 16.00 g/mol
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol
3. Next, we can calculate the number of moles of nitrogen and hydrogen:
Moles of nitrogen = Mass of nitrogen / Molar mass of nitrogen
Moles of nitrogen = 1.00 g / 14.01 g/mol ≈ 0.071 mol
Moles of hydrogen = Mass of hydrogen / Molar mass of hydrogen
Moles of hydrogen = 0.216 g / 1.01 g/mol ≈ 0.214 mol
4. Now, we can use the molar ratios to determine the amount of oxygen required:
Moles of oxygen needed = (Moles of nitrogen) * (Moles of oxygen / Moles of nitrogen)
Moles of oxygen needed = 0.071 mol * (1 mol/1 mol) = 0.071 mol
5. Finally, we can calculate the weight of oxygen required:
Weight of oxygen = Moles of oxygen needed * Molar mass of oxygen
Weight of oxygen = 0.071 mol * 16.00 g/mol ≈ 1.14 g
Therefore, to combine with 1.00 g of nitrogen to form an oxide, approximately 1.14 grams of oxygen would be required.
Which oxide? Any one of the several that are known? I will assume NO.
Note the following:
For H2O.
(0.126g H2/2 H atoms) x 16 = 1.00 g oxygen
and
(0.216g H2/3 H atoms) x 14 = 1.00 g N, so
(1.00/14) x 16 = ? g N to form NO.