Ammonia is produced when Nitrogen gas reacts with Hydrogen gas, as shown by the equation below:
N2 + 3H2 --> 2NH3
If the percentage yield is 90%, what would be your experimental yield of NH3 if 145.8g of N2 react?
2) 2Mg + O2 --> 2MgO
A) If 50g of Mg reacts with 50g of O2, what is the limiting reactants?
B) How much excess reactant is left over
C) How much product is produced?
mols N2 = 145.8g/molar mass N2 = ?
Convert mols N2 to mols NH3 using the coefficients in the balanced equation.
Now convert mols NH3 to g. g = mols x molar mass. This is the theoretical yield. If the actual yield is only 90%, then g NH3 x 0.90 = ?
To calculate the experimental yield of NH3 (ammonia) in the first equation and determine the limiting reactant and excess reactant, as well as the amount of product produced in the second equation, we need to follow a few steps.
1) Determine the molar masses of the compounds involved in the reactions:
- N2 (Nitrogen gas) has a molar mass of 28.02 g/mol
- H2 (Hydrogen gas) has a molar mass of 2.02 g/mol
- NH3 (Ammonia) has a molar mass of 17.03 g/mol
2) Calculate the moles of N2 using the given mass (145.8g) and its molar mass:
moles of N2 = mass of N2 / molar mass of N2
moles of N2 = 145.8g / 28.02 g/mol = 5.20 mol
3) Determine the theoretical yield of NH3 by using the stoichiometric ratio from the balanced equation (2 moles of NH3 produced per mole of N2 reacted):
theoretical yield of NH3 = moles of N2 x (2 moles of NH3 / 1 mole of N2)
theoretical yield of NH3 = 5.20 mol x (2 mol NH3 / 1 mol N2) = 10.40 mol NH3
4) Calculate the experimental yield of NH3 using the given percentage yield (90%):
experimental yield of NH3 = theoretical yield of NH3 x (percentage yield / 100)
experimental yield of NH3 = 10.40 mol x (90/100) = 9.36 mol NH3
Therefore, the experimental yield of NH3 from the given reaction would be 9.36 mol.
Moving on to the second equation:
A) To determine the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation (2 moles of Mg per mole of O2).
Calculate the moles of Mg and O2 using their given masses:
moles of Mg = mass of Mg / molar mass of Mg
moles of O2 = mass of O2 / molar mass of O2
moles of Mg = 50g / 24.31 g/mol = 2.06 mol
moles of O2 = 50g / 32.00 g/mol = 1.56 mol
Since the stoichiometric ratio requires 2 moles of Mg for every mole of O2, Mg is the limiting reactant because we have only 2.06 moles of Mg available compared to 1.56 moles of O2.
B) To find the excess reactant, subtract the number of moles needed for the limiting reactant from the actual number of moles present:
excess moles of Mg = moles of Mg - (moles of O2 x (2 moles of Mg / 1 mole of O2))
excess moles of Mg = 2.06 mol - (1.56 mol x (2 mol Mg / 1 mol O2)) = 2.06 mol - 3.12 mol = -1.06 mol
Since the excess moles of Mg are negative, we don't have any excess Mg left over. It means that all the Mg reacts with O2.
C) The amount of product produced can be calculated using the stoichiometric ratio from the balanced equation (2 moles of MgO produced per 1 mole of Mg reacted):
moles of MgO = moles of Mg x (2 moles of MgO / 2 moles of Mg)
moles of MgO = 2.06 mol x (2 mol MgO / 2 mol Mg) = 2.06 mol
To convert moles of MgO to grams, use its molar mass:
mass of MgO = moles of MgO x molar mass of MgO
mass of MgO = 2.06 mol x 40.31 g/mol = 83.08 g
Therefore, the amount of MgO produced is 83.08 grams.