Nitrogen reacts with hydrogen to make ammonia: N2 + 3H2 -> 2NH3
Calculate the maximum thoeretical yield of ammonia that can be made by reacting 90g of hyrdrogen with an excess of nitrogen.
Thanks!
To calculate the maximum theoretical yield of ammonia, we need to determine the limiting reagent between nitrogen and hydrogen. The limiting reagent is the reactant that gets completely consumed during the reaction, thereby limiting the amount of product that can be formed.
First, let's calculate the molar mass of hydrogen (H2) and nitrogen (N2):
- Hydrogen (H2): 2 (atomic mass of hydrogen) = 2 g/mol
- Nitrogen (N2): 2 (atomic mass of nitrogen) = 28 g/mol
Next, we calculate the number of moles of hydrogen (H2) using the given mass (90 g) and its molar mass (2 g/mol):
Number of moles = Mass / Molar mass
Number of moles of H2 = 90 g / 2 g/mol = 45 mol
Since hydrogen (H2) is in excess, we can use the stoichiometry of the balanced equation to find the maximum theoretical yield of ammonia (NH3). According to the balanced equation, 3 moles of hydrogen are required to react with 1 mole of nitrogen to form 2 moles of ammonia.
Therefore, the maximum theoretical yield of ammonia (NH3) can be calculated as follows:
Maximum theoretical yield = (Number of moles of H2) x (2 moles of NH3 / 3 moles of H2)
Maximum theoretical yield = (45 mol) x (2 mol NH3/ 3 mol H2)
Maximum theoretical yield = 30 mol NH3
Finally, to convert the moles of ammonia to grams, we multiply the number of moles by the molar mass of ammonia (2 x atomic mass of nitrogen + 3 x atomic mass of hydrogen):
Mass of NH3 = Number of moles x Molar mass
Mass of NH3 = 30 mol x (2 x 14 g/mol + 3 x 2 g/mol)
Mass of NH3 = 30 mol x (28 g/mol + 6 g/mol)
Mass of NH3 = 30 mol x 34 g/mol
Mass of NH3 = 1020 g
Therefore, the maximum theoretical yield of ammonia that can be produced is 1020 grams.
To calculate the maximum theoretical yield of ammonia, you need to determine which reactant limits the reaction. In this case, we are given an excess of nitrogen, meaning that hydrogen will be the limiting reactant.
To find the limiting reactant, you need to convert the given mass of hydrogen to moles. The molar mass of hydrogen (H2) is 2 grams/mole.
First, calculate the number of moles of hydrogen:
90 grams H2 x (1 mole H2 / 2 grams H2) = 45 moles H2
Now, we will use the balanced equation to determine the stoichiometry of the reaction. From the equation:
N2 + 3H2 -> 2NH3
According to the stoichiometry, 3 moles of hydrogen react with 1 mole of nitrogen to produce 2 moles of ammonia.
Since we have 45 moles of hydrogen, we can determine how many moles of ammonia can be produced:
45 moles H2 x (2 moles NH3 / 3 moles H2) = 30 moles NH3
Finally, to convert the moles of ammonia to grams, you multiply by the molar mass of ammonia. The molar mass of ammonia (NH3) is 17 grams/mole.
So, the maximum theoretical yield of ammonia is:
30 moles NH3 x (17 grams NH3 / 1 mole NH3) = 510 grams NH3
Therefore, the maximum theoretical yield of ammonia that can be made by reacting 90 grams of hydrogen with an excess of nitrogen is 510 grams NH3.
Write the equation and balance it.
Convert 90 g H2 to mols. mols = g/molar mass.
Convert mols H2 to mols NH3 by using the coefficients in the balanced equation.
Convert mols NH3 to grams NH3.
g NH3 = mols NH3 x molar mass NH3. That is the theoretical yield.