Ammonia reacts with oxygen to form nitrogen dioxide and water according to the following equation: 4NH3(g) + 7O2(g) ® 4NO2(g) + 6H2O(g)

You react ammonia and oxygen, and at the end of the experiment you find that you produced 27.0 g of water, and have 8.52 g of ammonia left over. Determine the mass of oxygen reacted (3 points)

To solve this problem, we need to determine the mass of oxygen reacted.

Let's start by finding the molar mass of ammonia (NH₃) and water (H₂O).

The molar mass of ammonia (NH₃) can be calculated as:
Molar mass of nitrogen (N) = 14.01 g/mol
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of ammonia (NH₃) = (1 * Molar mass of Nitrogen) + (3 * Molar mass of Hydrogen)
= (1 * 14.01) + (3 * 1.01)
= 14.01 + 3.03
= 17.04 g/mol

The molar mass of water (H₂O) can be calculated as:
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Molar mass of water (H₂O) = (2 * Molar mass of Hydrogen) + Molar mass of Oxygen
= (2 * 1.01) + 16.00
= 2.02 + 16.00
= 18.02 g/mol

Now, we can calculate the number of moles of water produced:
Given mass of water produced = 27.0 g
Number of moles of water = Mass of water / Molar mass of water
= 27.0 / 18.02
= 1.498 moles

The stoichiometry of the balanced equation tells us that for every 4 moles of NH₃, we produce 6 moles of H₂O. Therefore, the number of moles of NH₃ reacted can be calculated as:
Number of moles of NH₃ reacted = (Number of moles of water produced * 4) / 6
= (1.498 * 4) / 6
= 0.9987 moles

Now, let's calculate the mass of oxygen reacted:
Given mass of ammonia left over = 8.52 g
Number of moles of ammonia left over = Mass of ammonia / Molar mass of ammonia
= 8.52 / 17.04
= 0.5 moles

Number of moles of oxygen reacted = 7 * (Number of moles of ammonia reacted)
= 7 * 0.9987
= 6.99 moles

Now, we can calculate the mass of oxygen reacted:
Mass of oxygen reacted = Number of moles of oxygen reacted * Molar mass of oxygen
= 6.99 * (16.00 g/mol)
= 111.84 g

Therefore, the mass of oxygen reacted is approximately 111.84 g.

To determine the mass of oxygen reacted, we can start by calculating the molar mass of water, NH3, and NO2 using the periodic table.

The molar mass of water (H2O) is the sum of the molar masses of two hydrogen atoms (2 x 1.01 g/mol) and one oxygen atom (16.00 g/mol). Therefore, the molar mass of water is 18.02 g/mol.

The molar mass of ammonia (NH3) is the sum of the molar mass of one nitrogen atom (14.01 g/mol) and three hydrogen atoms (3 x 1.01 g/mol). Therefore, the molar mass of ammonia is 17.03 g/mol.

The molar mass of nitrogen dioxide (NO2) is the sum of the molar mass of one nitrogen atom (14.01 g/mol) and two oxygen atoms (2 x 16.00 g/mol). Therefore, the molar mass of nitrogen dioxide is 46.01 g/mol.

Now, let's calculate the number of moles of water produced and the number of moles of ammonia reacted.

The number of moles of water can be calculated using the formula:
Number of moles = mass / molar mass.
Number of moles of water = 27.0 g / 18.02 g/mol = 1.5 moles.

The number of moles of ammonia can be calculated using the same formula:
Number of moles = mass / molar mass.
Number of moles of ammonia = 8.52 g / 17.03 g/mol = 0.5 moles.

Now let's determine the stoichiometric ratio between water and oxygen in the balanced chemical equation.

From the equation 4NH3(g) + 7O2(g) -> 4NO2(g) + 6H2O(g), we can see that for every 7 moles of oxygen, 6 moles of water are produced.

Using this ratio, we can calculate the number of moles of oxygen reacted:
Number of moles of oxygen = (Number of moles of water / 6) x 7
Number of moles of oxygen = (1.5 moles / 6) x 7 = 1.75 moles.

Finally, we can calculate the mass of oxygen reacted using the formula:
Mass = number of moles x molar mass.
Mass of oxygen = 1.75 moles x 32.00 g/mol (molar mass of oxygen)
Mass of oxygen = 56.0 g.

Therefore, the mass of oxygen reacted is 56.0 g.

71.7 L

Convert 27.0 g H2O to moles. moles = g/molar mass.

Using the coefficients in the balanced equation, convert moles H2O to moles O2.
Now convert mols O2 to grams. grams O2 = moles O2 x molar mass O2.