Please show work
1. N2 (g) + 3H2(g) ---> 2NH3(g)
100.0 g of nitrogen reacts completely with excess hydrogen, and 34.0 g of NH3 are obtained what is the percent yield of ammonia?
2. 2Al (s) + 3Cl2(g) ---> 2AlCl3(s)
Assume that 0.40 g of Al is mixed with 0.60 g Cl2.
(a) What is the limiting reactant?
(b) What is the maximum amount of AlCl3, in grams, that can be produced?
1. find how many moles of N2 is 100g. Then you should get twice that amount of moles of ammonia. Convert that to grams.
percent yield=massgot/masshould
2. figure the moles of aluminum, and moles of chlorine. for each mole of aluminum, you should have 1.5 moles of chlorine. If you have less chlorine than that, chlorine is limiting, if you have more chlorine than that, aluminum is limiting.
for the product, use the limiting reactant. If aluminum, you should get the same number of moles of product.
Determine theoretical yield.
mols N2 = 100 g/molar mass = ?
Convert mol N2 to mols NH3 using the coefficients in the balanced equation.
Now convert mols NH3 to grams. g = mols x molar mass. This is the theoretical yield.
%yield = (actual yield/theor yield)*100 = ?
2.
mols Al = grams/molar mass = about 0.016
mols Cl2 = grams/molar mass = about 0.008.
Using the coefficients in the balanced equation, convert mols Al to mols AlCl3.
That's 0.016 x (2 mol AlCl3/2 mol Al) = 0.016 x (2/2) = about 0.016 mol AlCl3.
Use the same process and convert mols Cl2 to mols AlCl3. That will be
0.008 mols Cl2 x (2 mol AlCl3/3 mol Cl2) = 0.008 x 2/3 = about 0.005.
The two answers for mol AlCl3 do not agree; obviously one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. That's the a part.
b. Using the smaller value for mols AlCl3, convert to grams. g = mols x molar mass.
1. To find the percent yield of ammonia, we need to compare the actual yield (the amount of NH3 obtained) to the theoretical yield (the amount of NH3 predicted by stoichiometry).
First, let's calculate the theoretical yield of NH3:
- Use the given mass of nitrogen to calculate the moles of nitrogen:
Molar mass of N2 = 14.01 g/mol + 14.01 g/mol = 28.02 g/mol
Moles of N2 = 100.0 g / 28.02 g/mol = 3.571 mol
- According to the balanced chemical equation, the mole ratio of N2 to NH3 is 1:2. Thus, the moles of NH3 will be twice the moles of N2:
Moles of NH3 = 2 * 3.571 mol = 7.142 mol
- Convert the moles of NH3 to grams using the molar mass of NH3:
Molar mass of NH3 = 14.01 g/mol + 1.01 g/mol * 3 = 17.04 g/mol
Theoretical yield of NH3 = 7.142 mol * 17.04 g/mol = 121.429 g
Now, let's calculate the percent yield:
Actual yield = 34.0 g
Percent yield = (Actual yield / Theoretical yield) * 100% = (34.0 g / 121.429 g) * 100% ≈ 27.98%
Therefore, the percent yield of ammonia is approximately 27.98%.
2. (a) To determine the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation.
- Calculate the number of moles of Al:
Molar mass of Al = 26.98 g/mol
Moles of Al = 0.40 g / 26.98 g/mol ≈ 0.0148 mol
- Calculate the number of moles of Cl2:
Molar mass of Cl2 = 2 * 35.45 g/mol = 70.90 g/mol
Moles of Cl2 = 0.60 g / 70.90 g/mol ≈ 0.0085 mol
- According to the balanced chemical equation, the mole ratio of Al to Cl2 is 2:3. Thus, the mole ratio of Al to AlCl3 is also 2:3. Therefore, to react with 0.0085 mol of Cl2, we would need (2/3) * 0.0085 mol ≈ 0.0057 mol of Al.
Comparing the moles of Al (0.0148 mol) to the moles of Al needed (0.0057 mol), we can see that Al is in excess, and Cl2 is the limiting reactant.
(b) To calculate the maximum amount of AlCl3 that can be produced, we need to find the stoichiometric ratio of AlCl3 to Cl2 in the balanced chemical equation.
- According to the balanced chemical equation, the mole ratio of AlCl3 to Cl2 is 2:3. Therefore, the number of moles of AlCl3 that can be produced is given by:
Moles of AlCl3 = (2/3) * 0.0085 mol = 0.0057 mol
- Convert the moles of AlCl3 to grams using the molar mass of AlCl3:
Molar mass of AlCl3 = 26.98 g/mol + 35.45 g/mol * 3 = 133.33 g/mol
Maximum amount of AlCl3 = 0.0057 mol * 133.33 g/mol ≈ 0.761 g
Therefore, the maximum amount of AlCl3 that can be produced is approximately 0.761 grams.