Mass of oxygen needed to react completely with 64.3 g ammonia?
4NH3 + 5O2 ------> 4NO + 6H2O
Convert g NH3 to mols. mols = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NH3 to mols O2. You see 4 mols NH3 will need 5 mols O2.
Convert mols O2 to grams O2. grams = mols x molar mass = ?
To determine the mass of oxygen needed to react completely with ammonia, you need to first balance the chemical equation for the reaction between ammonia (NH3) and oxygen (O2).
The balanced chemical equation is:
4NH3 + 5O2 --> 4NO + 6H2O
From the balanced equation, you can see that 4 moles of NH3 reacts with 5 moles of O2.
Now let's calculate the molar mass of ammonia (NH3) and oxygen (O2):
- Molar mass of NH3 = 1.01 g/mol (atomic mass of N) + 3(1.01 g/mol) (atomic mass of H) = 17.03 g/mol.
- Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol.
Next, calculate the number of moles of ammonia present in 64.3 g of ammonia using the formula:
moles = mass / molar mass
moles of NH3 = 64.3 g / 17.03 g/mol = 3.78 mol (rounded to two decimal places).
From the balanced equation, you know that for every 4 moles of NH3, you require 5 moles of O2. Therefore, you can set up a ratio:
5 moles O2 / 4 moles NH3
Now, calculate the moles of oxygen required by multiplying the moles of ammonia by the ratio:
moles of O2 = (5/4) x 3.78 mol = 4.72 mol (rounded to two decimal places).
Finally, calculate the mass of oxygen by multiplying the moles of O2 by its molar mass:
mass of O2 = 4.72 mol x 32.00 g/mol = 150.97 g.
Therefore, the mass of oxygen needed to react completely with 64.3 g of ammonia is approximately 150.97 g.