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??
Limiting reagent problems, basically, are two stoichiometry problems in one.
There is a "little" shorter method but I like the following one.
1. You have the equation.
2. Convert 18.1 g NH3 to moles. moles = grams/molar mass.
3. Convert 90.4 g CuO to moles the same way.
4. Using the coefficients in the balanced equation, convert moles NH3 to moles N2.
5. Same process, convert moles CuO to moles N2.
6. It is more than likely that the answers from the last two steps will not agree which means one of the answers for moles N2 is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
7. Using the value from the last step (the smaller one), convert moles to grams. g = moles x molar mass. This is the theoretical yield.
8. percent yield = (actual yield/theoretical yield)*100 = ??
To determine the limiting reagent, we need to calculate the moles of each reactant.
1. Calculate the number of moles for NH3:
Molar mass of NH3 = 14.007 grams/mol (N) + 1.008 grams/mol (H) x 3 = 17.031 grams/mol
Moles of NH3 = 18.1 grams / 17.031 grams/mol = 1.064 mol
2. Calculate the number of moles for CuO:
Molar mass of CuO = 63.546 grams/mol (Cu) + 15.999 grams/mol (O) = 79.545 grams/mol
Moles of CuO = 90.4 grams / 79.545 grams/mol = 1.137 mol
Next, we compare the moles of each reactant based on the balanced chemical equation:
2NH3(g) + 3CuO(s) = N2(g) + 3Cu(s) + 3H2O(g)
Ratio of NH3 to CuO = 2:3
Moles of NH3 / Ratio = 1.064 mol / 2 = 0.532 mol
Moles of CuO / Ratio = 1.137 mol / 3 = 0.379 mol
The limiting reagent is the one with the lower number of moles. In this case, CuO is the limiting reagent because it has fewer moles available for the reaction.
To calculate the theoretical yield of nitrogen gas (N2):
Moles of N2 formed = Moles of limiting reagent * Ratio of N2 to limiting reagent
Moles of N2 formed = 0.379 mol * (1 mol N2 / 3 mol CuO) = 0.126 mol N2
To find the mass of N2 in grams, we use the molar mass of N2:
Molar mass of N2 = 14.007 grams/mol (N) x 2 = 28.014 grams/mol
Mass of N2 formed = Moles of N2 formed * Molar mass of N2
Mass of N2 formed = 0.126 mol * 28.014 grams/mol = 3.525 g N2
Therefore, the theoretical yield of nitrogen gas (N2) is 3.525 grams.
To calculate the percentage yield of nitrogen gas:
Percentage yield = (Actual yield / Theoretical yield) * 100%
Percentage yield = (8.52 g / 3.525 g) * 100% = 241.84%
The percentage yield of nitrogen gas is 241.84%.
To determine the limiting reagent, we need to compare the number of moles of each reactant. The reactant that produces the least amount of product is the limiting reagent.
First, we need to convert the given masses of ammonia (NH3) and copper(II) oxide (CuO) into moles.
The molar mass of ammonia (NH3) is:
N: 14.01 g/mol
H: 1.01 g/mol (3 hydrogen atoms in NH3)
So, the molar mass of NH3 is:
14.01 g/mol + (3 * 1.01 g/mol) = 17.03 g/mol
To find the moles of NH3, we divide the given mass by the molar mass:
moles of NH3 = 18.1 g / 17.03 g/mol ≈ 1.063 mol
The molar mass of copper(II) oxide (CuO) is:
Cu: 63.55 g/mol
O: 16.00 g/mol
So, the molar mass of CuO is:
63.55 g/mol + 16.00 g/mol = 79.55 g/mol
To find the moles of CuO, we divide the given mass by the molar mass:
moles of CuO = 90.4 g / 79.55 g/mol ≈ 1.137 mol
According to the balanced chemical equation, the stoichiometric ratio between NH3 and CuO is 2:3. This means that for every 2 moles of NH3, we need 3 moles of CuO.
From the balanced equation:
2 moles of NH3 react with 3 moles of CuO.
Comparing the moles of NH3 and CuO, we can see that there is an excess of CuO.
1.063 mol of NH3 is equivalent to:
1.063 mol NH3 * (3 mol CuO / 2 mol NH3) = 1.5945 mol CuO
Since we only have 1.137 mol of CuO, which is less than 1.5945 mol, CuO is the limiting reagent.
To find the theoretical yield of nitrogen gas (N2), we need to determine the amount of N2 produced from the limiting reagent.
From the balanced chemical equation:
2 moles of NH3 produce 1 mole of N2.
Since we have determined that CuO is the limiting reagent, we can use the stoichiometric ratio between NH3 and N2 to calculate the amount of N2 produced.
1.063 mol of NH3 is equivalent to:
1.063 mol NH3 * (1 mol N2 / 2 mol NH3) = 0.5315 mol N2
Therefore, the theoretical yield of N2 is 0.5315 mol.
Now, let's calculate the percent yield of nitrogen gas (N2):
Percent yield = (Actual yield / Theoretical yield) * 100
Given:
Actual yield of N2 = 8.52 g
To find the moles of N2, we divide the given mass by the molar mass:
moles of N2 = 8.52 g / 28.02 g/mol ≈ 0.304 mol
Percent yield = (0.304 mol / 0.5315 mol) * 100 ≈ 57.22%
Therefore, the percent yield of nitrogen gas is approximately 57.22%.