When ammonia was reacted to copper(ll)oxide at a high temperature, it will produce nitrogen gas, copper metal and water vapour. an experiment was done with 18.0g of ammonia was reacted with 90.0g of copper(ll)oxide.
(a)Write balance chemical equations for this reaction.
(b)Determine the limiting reactant.
(c)Determine excess reactant
(d)Calculate mass of nitrogen gas obtain
(e)Calculate percentage yield if only 10.0g of nitrogen gas obtain.
a.
2NH3 + 3CuO ==> N2 + 3Cu + 3H2O
First this is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
b.
mols NH3 = grams/molar mass = ?
mols CuO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NH3 to mols N2.
Do the same and convert mols CuO to mols N2.
You will get different values for mols N2 so one of those must be wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that smaller value is the LR.
c.
Obviously the OTHER reactant is the excess reagent.
d.
grams = mols x molar mass = ? This is the theoretical yield (TY).
e.
The actual yield (AY) is in the problem at 10.0 grams.
%yield = (AY/TY)*100 = ?
(a) The balanced chemical equation for the reaction can be written as:
4NH3 + 3CuO -> 2N2 + 3Cu + 6H2O
(b) To determine the limiting reactant, we need to compare the amount of product that can be formed from each reactant.
First, we find the moles of ammonia (NH3):
Molar mass of NH3 = 14.01 g/mol
moles of NH3 = mass/molar mass = 18.0 g / 14.01 g/mol ≈ 1.285 mol
Next, we find the moles of copper(II) oxide (CuO):
Molar mass of CuO = 79.55 g/mol
moles of CuO = mass/molar mass = 90.0 g / 79.55 g/mol ≈ 1.131 mol
From the balanced chemical equation, we can see that 4 moles of NH3 react with 3 moles of CuO to produce 2 moles of N2. Therefore, the stoichiometry ratio between NH3 and N2 is 4:2, or simplified as 2:1.
Since the ratio is 2:1, we can see that we have an excess of NH3 because 1.285 mol of NH3 would only react with 0.6425 mol of CuO (half the amount needed for stoichiometry).
(c) The excess reactant is ammonia (NH3).
(d) To calculate the mass of nitrogen gas obtained, we first need to find the moles of nitrogen gas (N2), using the stoichiometry of the balanced chemical equation. From the equation, we can see that 4 moles of NH3 produce 2 moles of N2.
moles of N2 = (1.285 mol NH3) x (2 mol N2/4 mol NH3) = 0.6425 mol N2
Next, we find the molar mass of N2:
molar mass of N2 = 28.02 g/mol
mass of N2 = moles x molar mass = 0.6425 mol x 28.02 g/mol ≈ 18.03 g
Therefore, the mass of nitrogen gas obtained is approximately 18.03 g.
(e) To calculate the percentage yield, we need to compare the actual yield (10.0 g) with the theoretical yield (calculated in part (d), which is 18.03 g).
percentage yield = (actual yield / theoretical yield) x 100%
percentage yield = (10.0 g / 18.03 g) x 100% ≈ 55.4%
Therefore, the percentage yield is approximately 55.4%.
(a) The balanced chemical equation for the reaction can be written as follows:
3CuO + 2NH3 -> 3Cu + 3H2O + N2
(b) To determine the limiting reactant, we need to calculate the amount of product that can be formed from each reactant. The balanced equation shows that for every 3 moles of CuO reacted, we will produce 1 mole of N2. Similarly, for every 2 moles of NH3 reacted, we also produce 1 mole of N2.
First, let's calculate the number of moles for each reactant:
- Moles of CuO = mass of CuO / molar mass of CuO
= 90.0g / 79.55 g/mol
≈ 1.13 mol
- Moles of NH3 = mass of NH3 / molar mass of NH3
= 18.0g / 17.03 g/mol
≈ 1.06 mol
Since the stoichiometric ratio of CuO to N2 is 3:1, it means that 1.13 moles of CuO will produce (1.13/3) ≈ 0.38 moles of N2. Whereas, the stoichiometric ratio of NH3 to N2 is 2:1, so 1.06 moles of NH3 will produce (1.06/2) ≈ 0.53 moles of N2.
Comparing these values, we find that the limiting reactant is CuO because it produces a smaller amount of moles of N2 compared to NH3.
(c) To determine the excess reactant, we calculated the amount of N2 produced from each reactant. Since CuO is the limiting reactant, it will be fully consumed, and the remaining reactant, NH3, will be in excess.
(d) To calculate the mass of N2 obtained, we need to use the stoichiometry of the reaction. From the balanced equation, we know that 3 moles of CuO produce 1 mole of N2. Therefore, the number of moles of N2 can be calculated as follows:
Moles of N2 = (moles of CuO) / 3
= 1.13 mol / 3
≈ 0.38 mol
Mass of N2 = (moles of N2) × (molar mass of N2)
= 0.38 mol × 28.01 g/mol
≈ 10.64 g
Thus, the mass of nitrogen gas obtained is approximately 10.64 g.
(e) To calculate the percentage yield, we need to compare the actual yield to the theoretical yield. The theoretical yield is the amount of product that should be obtained based on the stoichiometry of the balanced equation.
In this case, the actual yield is given as 10.0 g. The theoretical yield can be calculated using the stoichiometry of the balanced equation. We know that 3 moles of CuO produce 1 mole of N2, which corresponds to a reaction yield of 28.01 g.
Theoretical yield of N2 = (moles of CuO) / 3 × (molar mass of N2)
= 1.13 mol / 3 × 28.01 g/mol
≈ 9.37 g
Percentage yield = (actual yield / theoretical yield) × 100
= (10.0 g / 9.37 g) × 100
≈ 106.8%
Therefore, the percentage yield is approximately 106.8%.