consider the following balance chemical equation:
2NH3(g)+3CuO(s)=N2(g)+3Cu(s)+3H2O(g)
if 18.1grams of ammonia reacts with 90.4 grams of copper(II) oxide, which is the limitting reagent?? what is the theorectical yield of nitrogen gasd? if 8.52 g of N2 are formed, what is the % yield of nitrogn gas??
please check if is correct....
limitting reagent is NH3 with 2.13 mol N2
theoretical yield nitrogen gas is 59.68g N2
percent yield of nitrogen gas is 14.28%
please i need to know what i'm doing wrong..thank you....
To determine the limiting reagent and the theoretical yield of nitrogen gas, you need to follow these steps:
Step 1: Convert the given masses of ammonia (NH3) and copper (II) oxide (CuO) to moles.
Given:
Mass of NH3 = 18.1 g
Mass of CuO = 90.4 g
Molar masses:
NH3 = 17.03 g/mol
CuO = 79.55 g/mol
Moles of NH3 = Mass of NH3 / Molar mass of NH3 = 18.1 g / 17.03 g/mol ≈ 1.063 mol
Moles of CuO = Mass of CuO / Molar mass of CuO = 90.4 g / 79.55 g/mol ≈ 1.137 mol
Step 2: Use the balanced equation to determine the stoichiometry of the reaction.
From the balanced equation:
2NH3 + 3CuO = N2 + 3Cu + 3H2O
It is clear that the stoichiometric ratio between NH3 and N2 is 2:1. This means that for every 2 moles of NH3, we should form 1 mole of N2.
Step 3: Determine the limiting reagent.
To find the limiting reagent, we compare the stoichiometric ratio of the reactants with the actual ratio given by their moles.
For NH3: 2 moles of NH3 should produce 1 mole of N2
Actual ratio: 1.063 mol NH3 / 2 ≈ 0.5315 mol N2
For CuO: 3 moles of CuO should produce 1 mole of N2
Actual ratio: 1.137 mol CuO / 3 ≈ 0.379 mol N2
Comparing the two ratios, it is clear that the reactant with the smaller actual ratio is the limiting reagent. In this case, CuO has the smaller ratio of 0.379 mol N2, so it is the limiting reagent.
Step 4: Calculate the theoretical yield of nitrogen gas.
Since CuO is the limiting reagent, we use its stoichiometric ratio to calculate the theoretical yield of N2.
From the balanced equation:
2 moles of NH3 produce 1 mole of N2
Moles of NH3 = 1.063 mol
Moles of N2 = 1/2 * 1.063 mol = 0.5315 mol
Mass of N2 = Moles of N2 * Molar mass of N2 = 0.5315 mol * 28.02 g/mol ≈ 14.92 g
Therefore, the theoretical yield of N2 is approximately 14.92 grams.
To calculate the percentage yield of N2, use the given actual yield (8.52 g) and divide it by the theoretical yield (14.92 g). Then multiply by 100.
Percentage yield of N2 = (Actual yield / Theoretical yield) * 100
= (8.52 g / 14.92 g) * 100
≈ 57.12%
So, the correct answers are as follows:
- The limiting reagent is CuO.
- The theoretical yield of nitrogen gas is approximately 14.92 g.
- The percent yield of nitrogen gas is approximately 57.12%.
It seems that there was an error in your calculations for the theoretical yield and percentage yield. Please double-check your calculations or follow the steps I provided to get the correct answers.
To determine the limiting reagent, we need to calculate the number of moles of each reactant.
Given:
Mass of NH3 = 18.1 g
Molar mass of NH3 = 17.03 g/mol
Moles of NH3 = (Mass of NH3) / (Molar mass of NH3) = 18.1 g / 17.03 g/mol = 1.063 mol
Given:
Mass of CuO = 90.4 g
Molar mass of CuO = 79.55 g/mol
Moles of CuO = (Mass of CuO) / (Molar mass of CuO) = 90.4 g / 79.55 g/mol = 1.137 mol
According to the balanced chemical equation, the ratio of moles of NH3 to moles of CuO is 2:3. Therefore, for every 2 moles of NH3, we need 3 moles of CuO.
The stoichiometric ratio tells us that 1.063 mol of NH3 requires (3/2) x 1.063 mol = 1.595 mol of CuO.
Since we have 1.137 mol of CuO available, which is less than the required amount, CuO is the limiting reagent.
Now, let's calculate the theoretical yield of nitrogen gas (N2). The balanced equation tells us that from 2 moles of NH3, we can obtain 1 mole of N2.
Moles of N2 = 1.063 mol NH3 x (1 mol N2 / 2 mol NH3) = 0.5315 mol N2
The molar mass of N2 is 28.02 g/mol, so the theoretical yield of N2 is:
Theoretical yield of N2 = (Moles of N2) x (Molar mass of N2) = 0.5315 mol x 28.02 g/mol = 14.898 g of N2
Therefore, the theoretical yield of nitrogen gas is 14.898 grams.
To calculate the percent yield of nitrogen gas, we use the actual yield (8.52 g) and the theoretical yield (14.898 g):
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Percent Yield = (8.52 g / 14.898 g) x 100 = 57.16%
Therefore, the percent yield of nitrogen gas is 57.16%, not 14.28% as mentioned before.
I hope this clarifies everything. Let me know if you have any further questions!