2.00 g sample of ammonia is mixed with 4.00 g of oxygen. How much excess reactant remains after the reaction has stopped?
molar mass of NH3 = 14 + 3 = 17 grams/mol
so you have 2 grams ( 1 mol/17 g) = 0.118 mols of NH3
molar mass of O2 = 16*2 = 32 g /mol
so you have 4 g (1 mol/32 g) = 1/8 = 0.125 mols of O2
NH3 + O2 ----> H2O + N2 maybe? If N is not oxidized to NO2 or worse
Balance that
4 NH3 + 3 O2 ---> 6H2O + 2 N2
so for .118 mols of NH3 you need (3/4) mols of O2 = 0.0885 mols of O2
.0885 mols * 32 g/mol = 2.832 grams of O2 needed
but we have 4 grams of O2
so how much O2 is left over?
To find out how much of the excess reactant remains after the reaction has stopped, we first need to determine the limiting reactant.
To do this, we need to compare the number of moles of each reactant.
1. First, let's find the number of moles for each reactant:
- The molar mass of ammonia (NH3) is 17.03 g/mol.
- The molar mass of oxygen (O2) is 32.00 g/mol.
Number of moles of ammonia:
moles of ammonia = mass of ammonia / molar mass of ammonia
moles of ammonia = 2.00 g / 17.03 g/mol
Number of moles of oxygen:
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 4.00 g / 32.00 g/mol
2. Now, we compare the moles of ammonia and oxygen to determine the limiting reactant.
The ratio of ammonia to oxygen in the balanced chemical equation is 4:5.
If the ratio of moles of ammonia to oxygen is less than 4:5, ammonia is the limiting reactant. Otherwise, oxygen is the limiting reactant.
moles of ammonia / moles of oxygen = (2.00 g / 17.03 g/mol) / (4.00 g / 32.00 g/mol)
3. Calculate the ratio:
moles of ammonia / moles of oxygen = (2.00 g / 17.03 g/mol) * (32.00 g/mol / 4.00 g)
4. Simplify the ratio:
moles of ammonia / moles of oxygen = 0.1881
Since the ratio of moles of ammonia to moles of oxygen is less than the stoichiometric ratio of 4:5, ammonia is the limiting reactant.
5. Now, we can calculate the amount of excess reactant remaining using the limiting reactant:
To find the moles of ammonia reacted, we use the stoichiometry from the balanced chemical equation.
From the balanced chemical equation:
4 NH3 + 5 O2 => 4 NO + 6 H2O
Since we have 2 moles of ammonia, it will react with (2/4 * 5) = 2.5 moles of oxygen.
Moles of oxygen remaining:
moles of oxygen remaining = moles of oxygen - moles of oxygen reacted
moles of oxygen remaining = (4.00 g / 32.00 g/mol) - 2.5 mol
6. Calculate the mass of oxygen remaining:
Mass of oxygen remaining = moles of oxygen remaining * molar mass of oxygen
Mass of oxygen remaining = (4.00 g / 32.00 g/mol) - 2.5 mol * 32.00 g/mol
Finally, calculating the mass of oxygen remaining will give us the amount of excess reactant remaining after the reaction has stopped.
To find out how much excess reactant remains, we first need to determine the limiting reactant, which is the reactant that is completely consumed first and determines the amount of product formed.
To find the limiting reactant, we compare the amount of moles of each reactant. The balanced chemical equation for the reaction is:
4NH3 + 5O2 -> 4NO + 6H2O
1. Convert the masses of both reactants into moles using their molar masses:
- Ammonia (NH3): Molar mass = 14.01 g/mol
Moles of NH3 = 2.00 g / (14.01 g/mol) = 0.14281 mol
- Oxygen (O2): Molar mass = 32.00 g/mol
Moles of O2 = 4.00 g / (32.00 g/mol) = 0.125 mol
2. Next, find the mole ratio of the reactants from the balanced equation.
According to the balanced equation, the mole ratio of NH3 to O2 is 4:5.
This means that for every 4 moles of NH3, 5 moles of O2 are required.
3. Determine which reactant is the limiting reactant.
To do this, compare the moles of each reactant to the stoichiometric ratio.
Moles of NH3 / Stoichiometric coefficient of NH3 = 0.14281 mol / 4 = 0.0357 mol
Moles of O2 / Stoichiometric coefficient of O2 = 0.125 mol / 5 = 0.025 mol
From the comparison, we can see that the moles of O2 (0.025 mol) are lower than the moles of NH3 (0.0357 mol). Therefore, O2 is the limiting reactant.
4. Calculate the moles of the excess reactant that remains.
Since O2 is the limiting reactant, all of it will be consumed in the reaction. This means that the excess reactant is NH3.
Excess moles of NH3 = Moles of NH3 - Moles of O2
Excess moles of NH3 = 0.0357 mol - 0.025 mol = 0.0107 mol
5. Convert the excess moles of NH3 back into grams:
Mass of NH3 = Excess moles of NH3 * Molar mass of NH3
Mass of NH3 = 0.0107 mol * 14.01 g/mol = 0.150 g
Therefore, there will be approximately 0.150 grams of excess NH3 remaining after the reaction has stopped.