a mixture of 34.0g of ammonia and 50.0g of elemental oxygen react to form elemental nitrogen and water. Please write a balanced chemical equation for this reaction. Find the theoretical yield of water in grams and give the amount of excess reactant in grams
4NH3 + 3O2 ==> 2N2 + 6H2O
mols NH3 = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NH3 to mols H2O.
Do the same for mols O2 to mols H2O.
It is likely that these two values will not agree which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the regent producing that value is the limiting regent.
Using the smaller value, convert mols N2 to g N2. g = mols x molar mass = ?
After you know the identity of the limiting reagent (LR), repeat step 2 of converting mols or LR to mols of non-LR used. Then subtract from the initial value to find amount remaining in excess.
To write a balanced chemical equation for this reaction, we first need to determine the chemical formulas of the reactants and products involved.
Ammonia (NH3) is composed of one nitrogen atom (N) and three hydrogen atoms (H). Elemental oxygen (O2) consists of two oxygen atoms bonded together. The reaction forms elemental nitrogen (N2) and water (H2O).
The balanced chemical equation for this reaction can be written as follows:
4 NH3 + 5 O2 -> 4 N2 + 6 H2O
To find the theoretical yield of water in grams, we need to determine which reactant is limiting and which is in excess. This can be done by using the concept of stoichiometry and finding the moles of the reactants.
1. Convert the mass of each reactant to moles:
Mass of NH3 = 34.0g
Molar mass of NH3 = 17.03g/mol
Number of moles of NH3 = 34.0g / 17.03g/mol = 1.998 moles
Mass of O2 = 50.0g
Molar mass of O2 = 32.00g/mol
Number of moles of O2 = 50.0g / 32.00g/mol = 1.563 moles
2. Use the stoichiometry from the balanced equation to determine the moles of water produced:
From the balanced equation, the stoichiometric ratio of NH3 to H2O is 4:6.
Therefore, moles of H2O = 1.998 moles NH3 * (6 moles H2O / 4 moles NH3) = 2.997 moles H2O
3. Convert the moles of water to grams:
Molar mass of H2O = 18.015g/mol
Mass of water = 2.997 moles H2O * 18.015g/mol = 54.03g
Thus, the theoretical yield of water is 54.03 grams.
To find the amount of excess reactant, we compare the moles used in the reaction to the initial moles of the reactant:
Initial moles of O2 = 1.563 moles
Moles of O2 used in the reaction = 5 moles of O2 * (1.998 moles NH3 / 4 moles NH3) = 2.4975 moles
Excess moles of O2 = Initial moles of O2 - Moles of O2 used in the reaction = 1.563 moles - 2.4975 moles = -0.9345 moles
Since the excess moles of O2 is negative, it indicates that there is no excess O2 remaining. Thus, the amount of excess reactant in grams is 0 grams.