2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g)
a. What mass of acrylonitrile can be produced from a mixture of 1.24 kg of propylene (C3H6), 1.61 kg of ammonia, and 1.90 kg of oxygen, assuming 100% yield?
b. What mass of water is produced?
This is a limiting reagent problem. We know that because an amount is given for more than one of the reactants. This is a little more difficult than when only two reactants are present but I work these the same way.
Convert 1.24 kg propylene to mols. mols = grams/molar mass,
Convert 1.61 kg NH3 to mols.
Convert 1.90 kg O2 to mols.
Using the coefficients in the balanced equation convert mols propylene to mols acrylonitrile.
Do the same for NH3.
Do he same for O2.
It is likely that the values obtained for mols acrylonitrile will not agree which means two of them are wrong. The correct value in limiting reagent problems is ALWAYS the smallest value and the reagent producing that value is the limiting reagent (LR). Using that value, calculate grams acrylonitrile by g = mols x molar mass.
b.
Using the value of mols for limiting reagent, convert from mols LR to mols H2O. Then g H2O = mols H2O x molar mass H2O
To solve this problem, we will use the stoichiometry of the balanced chemical equation to calculate the masses involved.
Step 1: Convert the given masses to moles.
a. Propylene (C3H6):
Molar mass of C3H6 = (3 x 12.01 g/mol) + (6 x 1.01 g/mol) = 42.08 g/mol
Moles of C3H6 = mass (g) / molar mass (g/mol)
Moles of C3H6 = 1.24 kg x (1000 g/kg) / 42.08 g/mol = 29.47 mol
b. Ammonia (NH3):
Molar mass of NH3 = 14.01 g/mol + (3 x 1.01 g/mol) = 17.03 g/mol
Moles of NH3 = mass (g) / molar mass (g/mol)
Moles of NH3 = 1.61 kg x (1000 g/kg) / 17.03 g/mol = 94.67 mol
c. Oxygen (O2):
Molar mass of O2 = (2 x 16.00 g/mol) = 32.00 g/mol
Moles of O2 = mass (g) / molar mass (g/mol)
Moles of O2 = 1.90 kg x (1000 g/kg) / 32.00 g/mol = 59.38 mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we compare the mole ratios of the reactants from the balanced chemical equation.
From the balanced equation:
2 moles C3H6 produces 2 moles C3H3N
2 moles NH3 produces 2 moles C3H3N
3 moles O2 produces 2 moles C3H3N
The reactant that forms the least amount of C3H3N will be the limiting reactant.
Using the moles obtained in step 1, we can compare the mole ratios to determine the limiting reactant:
Moles of C3H6 = 29.47 mol
Moles of NH3 = 94.67 mol
Moles of O2 = 59.38 mol
The mole ratio of C3H6 to C3H3N is 2:2, which simplifies to 1:1.
The mole ratio of NH3 to C3H3N is 2:2, which simplifies to 1:1.
The mole ratio of O2 to C3H3N is 3:2.
Comparing the mole ratios, we can see that O2 has the smallest ratio. Therefore, O2 is the limiting reactant.
Step 3: Calculate the theoretical yield of acrylonitrile.
Using the mole ratios from the balanced equation, we can calculate the theoretical yield of acrylonitrile.
From the balanced equation:
2 moles C3H6 produces 2 moles C3H3N
Therefore, 29.47 mol of C3H6 will produce 29.47 mol of C3H3N.
Step 4: Convert the moles of C3H3N to the mass of acrylonitrile.
Molar mass of C3H3N = (3 x 12.01 g/mol) + (3 x 1.01 g/mol) + (1 x 14.01 g/mol) = 53.06 g/mol
Mass of acrylonitrile = moles of C3H3N x molar mass of C3H3N
Mass of acrylonitrile = 29.47 mol x 53.06 g/mol = 1,562.04 g
a. Therefore, the maximum mass of acrylonitrile that can be produced is 1,562.04 grams.
Step 5: Calculate the mass of water produced.
From the balanced equation:
2 moles C3H6 produces 6 moles H2O
Therefore, 29.47 mol of C3H6 will produce (6/2) x 29.47 = 88.41 mol of H2O.
Molar mass of H2O = 2 x 1.01 g/mol + 16.00 g/mol = 18.02 g/mol
Mass of water = moles of H2O x molar mass of H2O
Mass of water = 88.41 mol x 18.02 g/mol = 1592.04 g
b. Therefore, the maximum mass of water produced is 1592.04 grams.
To find the mass of acrylonitrile (C3H3N) that can be produced, we need to use the given information and calculate the limiting reactant. The limiting reactant is the reactant that is completely consumed in a reaction and determines the amount of product that can be formed.
a. First, let's calculate the number of moles of each reactant using their molar masses:
- Propylene (C3H6): Molar mass = 3(12.01 g/mol) + 6(1.01 g/mol) = 42.08 g/mol
Moles of propylene = mass of propylene / molar mass = 1.24 kg / 42.08 g/mol
- Ammonia (NH3): Molar mass = 1(14.01 g/mol) + 3(1.01 g/mol) = 17.03 g/mol
Moles of ammonia = mass of ammonia / molar mass = 1.61 kg / 17.03 g/mol
- Oxygen (O2): Molar mass = 2(16.00 g/mol) = 32.00 g/mol
Moles of oxygen = mass of oxygen / molar mass = 1.90 kg / 32.00 g/mol
b. Next, we need to compare the mole ratios of the reactants to determine the limiting reactant. According to the balanced chemical equation:
2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g)
The ratio between propylene and acrylonitrile is 2:2 (1:1), between ammonia and acrylonitrile is 2:2 (1:1), and between oxygen and acrylonitrile is 3:2.
To determine the limiting reactant, we compare the number of moles of each reactant to the reactant with the smallest stoichiometric coefficient. In this case, it is propylene (C3H6).
c. The limiting reactant is propylene (C3H6). Therefore, we need to calculate the moles of acrylonitrile produced from the moles of propylene using the stoichiometry ratio:
Moles of acrylonitrile = Moles of propylene
d. Finally, we can calculate the mass of acrylonitrile produced using the moles of acrylonitrile and its molar mass:
Mass of acrylonitrile = Moles of acrylonitrile × molar mass of acrylonitrile
b. The balanced chemical equation tells us that for every 2 moles of acrylonitrile produced, 6 moles of water are produced. Therefore, we can calculate the moles of water produced using the moles of acrylonitrile:
Moles of water = 3 × Moles of acrylonitrile
Finally, we can calculate the mass of water produced using the moles of water and its molar mass:
Mass of water = Moles of water × molar mass of water
By following these steps, you should be able to determine the masses of acrylonitrile and water produced.