Industrially, nitric acid is produced by the Ostwald process represented by the following equations:
4NH3 +5O2 = 4NO +H2O
2NO +O2 = 2NO2
2NO2 +H2O = HNO3 +HNO2
What mass of NH3 (in grams) must be used to produce 1.0 ton of HNO3 by the above procedure, assuming an 80 percent of yield in each step?(1ton=200 lb,1lb=453.6g)
Note the first equation isn't balanced. Note that 1 ton = 2000 lb. Note that the third equation isn't what I've seen for the Ostwald process but I've used what you posted.
You want 2000 lb x 453.6 g/lb x (1/0.8) x (1/0.8) x (1/0.8) = approx 2E6 g to be produced considering the 80% yield in each step but that's an estimate and you need a better answer than that.
So NH3 must be
2E6 g HNO3 x (2 mol NO21 mol HNO3) x (2 mols NO/2 mols NO2) x (4 molsNH3/4 mols NO) = ? g NH3.
Post your work if you get stuck.
Well, before I give you a truly explosive answer, let's do some calculations, shall we?
First, let's figure out how many grams are in 1 ton. We know that 1 ton is equal to 2000 pounds, and 1 pound is equal to 453.6 grams. So, one ton is equal to 2000 times 453.6 grams, which is 907,200 grams.
Now, let's assume 80 percent yield for each step. That means each step will produce 80 percent of the product that is expected. So, we need to take that into account.
In the first step, 4 moles of NH3 is needed to produce 4 moles of HNO3. So, we need the molar mass of NH3 to figure out how many grams are required. The molar mass of NH3 is approximately 17 grams per mole.
To find out how many moles of NH3 are needed, we can set up a ratio:
4 moles NH3 / 4 moles HNO3 = x grams NH3 / 1.0 ton HNO3
Simplifying that, we get:
x = (4 moles NH3 / 4 moles HNO3) * (1.0 ton HNO3 / 907,200 grams NH3)
Note that the ratio is 4 moles NH3 to 4 moles HNO3 because the coefficients in the balanced equation are the same.
Plugging in the numbers, we get:
x = (4 / 4) * (1.0 ton / 907,200 g) * (907,200 g / 1 ton)
The tons cancel out, and we're left with:
x = 1.0
So, we need 1.0 gram of NH3 to produce 1.0 ton of HNO3. But keep in mind, this is assuming 80 percent yield at each step. I hope this answer doesn't blow you away!
To solve this problem, we'll need to calculate the number of moles of NH3 required to produce 1.0 ton of HNO3.
Step 1: Convert the given mass of HNO3 from tons to grams.
1 ton = 2000 pounds (lb)
1 lb = 453.6 grams (g)
Therefore, 1 ton = 2000 lb * 453.6 g/lb = 907,200 grams.
Step 2: Calculate the number of moles of HNO3 produced from the given mass.
The molar mass of HNO3 is:
H (1.01 g/mol) + N (14.01 g/mol) + O (16.00 g/mol) = 63.02 g/mol.
Moles of HNO3 = mass of HNO3 / molar mass of HNO3
= 907,200 g / 63.02 g/mol
≈ 14,400 mol
Step 3: Calculate the number of moles of NH3 required based on the stoichiometry of the reactions.
From the balanced equations, we can see that 4 moles of NH3 are required to produce 4 moles of NO, and 4 moles of NO are required to produce 4 moles of HNO3.
Therefore, 1 mole of HNO3 requires 1 mole of NH3.
Step 4: Calculate the mass of NH3 required.
The molar mass of NH3 is:
N (14.01 g/mol) + H (1.01 g/mol x 3) = 17.03 g/mol
Moles of NH3 = moles of HNO3
= 14,400 mol
Mass of NH3 = moles of NH3 x molar mass of NH3
= 14,400 mol x 17.03 g/mol
≈ 245,472 g
Therefore, approximately 245,472 grams (or 245.47 kg) of NH3 must be used to produce 1.0 ton of HNO3 by the Ostwald process.
To find the mass of NH3 required to produce 1.0 ton of HNO3, we need to follow these steps:
Step 1: Convert the given weight of 1.0 ton to grams.
1 ton = 2000 pounds (since 1 ton = 2000 lb)
1 pound = 453.6 grams
1 ton = 2000 lb * 453.6 g/lb = 907,200 g
Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that 4 moles of NH3 react to produce 4 moles of NO, and then 2 moles of NO react to produce 1 mole of HNO3. Therefore, the molar ratio between NH3 and HNO3 is 4:1.
Step 3: Calculate the molar mass of NH3.
NH3 has a molar mass of:
(1 * atomic mass of N) + (3 * atomic mass of H)
= (1 * 14.01 g/mol) + (3 * 1.01 g/mol)
= 17.04 g/mol
Step 4: Calculate the number of moles of HNO3.
Using the molar mass of HNO3:
1 ton of HNO3 = 907,200 g
Molar mass of HNO3 = 63.01 g/mol
Number of moles of HNO3 = (mass of HNO3) / (molar mass of HNO3)
= 907,200 g / 63.01 g/mol
Step 5: Calculate the number of moles of NH3.
Since the molar ratio of NH3 to HNO3 is 4:1, the number of moles of NH3 is four times the number of moles of HNO3.
Number of moles of NH3 = (4 * number of moles of HNO3)
Step 6: Convert the number of moles of NH3 to grams.
Mass of NH3 = (number of moles of NH3) * (molar mass of NH3)
Finally, we can substitute the values into the equations to find the mass of NH3 required.
(Note: For simplicity, we will use the given yield of 80 percent for each step, but in practice, the process efficiency may vary.)
Let's calculate it step by step:
Step 1:
Weight of 1.0 ton = 907,200 grams
Step 2:
Molar ratio NH3:HNO3 = 4:1
Step 3:
Molar mass of NH3 = 17.04 g/mol
Step 4:
Number of moles of HNO3 = (907,200 g) / (63.01 g/mol)
Step 5:
Number of moles of NH3 = (4 * number of moles of HNO3)
Step 6:
Mass of NH3 = (number of moles of NH3) * (molar mass of NH3)
Substitute the values into the equations to find the mass of NH3 required.