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.
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?
2 answers
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.
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.