Nitric acid is produced commercially by the Ostwald
process, represented by the following equations:
4NH3(g) + 5O2(g) > 4NO(g) + 6H2O(g)
2NO(g) + O2(g) > 2NO2 (g)
3NO2(g) + H2O(l) > 2HNO3(aq) + NO(g)
What mass of NH3 must be used to produce 1.0x10^6 kg HNO3 by the Ostwald
process? Assume 100% yield in each reaction [and assume that the NO produced in the third step is not recycled.]
--HNO3 mols =15873015.87
which means 95238095.24 mols of NO2
Because the coefficient of NO2 and NO and NO and NH4 are the same, there are 95238095.24 NH4.
Multiply that by the molar mass (17) and divide by 1000 gets me 1619047.619 kg. Is this right?
No, I don't think it is right.
mols HNO3 is correct EXCEPT that you don't have that many significant figures.
1.0E6 kg x (1000g/kg)= 1.0E9 g
1.0E9g x 1 mol HNO3/63 g = 1.587E7 (and that is too many places, also).
I THINK you used the wrong factor to convert to mols NO2. I THINK you used
1.587E7 x 2/3 but it should be 3/2; i.e.,
mols HNO3 x (3 mols NO2/2 mols HNO3) = mols NO2. You are correct that everyting else is 1:1 so mols NO2 will be same as mols NH3 and you convert that to g NH3 by multiplying the molar mass NH3 as you posted. Check my thinking.
Apparently i did use some wrong factor.
Is the answer 4.0 x 10^5 kg correct?
Yes, that is correct.
To find the correct answer, let's calculate step by step.
Given:
Mass of HNO3 to be produced = 1.0x10^6 kg
First, let's convert the mass of HNO3 to moles:
1.0x10^6 kg * (1000 g/kg) = 1.0x10^9 g
1.0x10^9 g / 63 g/mol (molar mass of HNO3) = 1.5873x10^7 mol HNO3
Next, we need to calculate the moles of NH3 required, using the balanced equation:
4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)
Since the coefficients are 4:4, the moles of NH3 required will be the same as the moles of NO produced. Therefore, the moles of NH3 required will be 1.5873x10^7 mol.
Finally, convert moles of NH3 to mass:
1.5873x10^7 mol * 17 g/mol (molar mass of NH3) = 2.699x10^8 g
2.699x10^8 g / 1000 = 2.699x10^5 kg
Therefore, the correct answer is 2.699x10^5 kg of NH3 required to produce 1.0x10^6 kg of HNO3 by the Ostwald process.
To determine the mass of NH3 needed to produce 1.0x10^6 kg of HNO3 by the Ostwald process, we'll need to go step by step through the given equations.
1. Calculate the number of moles of HNO3 needed:
Mass of HNO3 = 1.0x10^6 kg = 1.0x10^9 g
Molar mass of HNO3 = 63 g/mol
Moles of HNO3 = Mass of HNO3 / Molar mass of HNO3
= 1.0x10^9 g / 63 g/mol
= 1.5873x10^7 mol
2. Calculate the moles of NO2 produced:
From the equation 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g),
The stoichiometric ratio between HNO3 and NO2 is 2:3.
Therefore, Moles of NO2 = Moles of HNO3 x (3 mol NO2 / 2 mol HNO3)
= 1.5873x10^7 mol x (3 / 2)
= 2.38095x10^7 mol
3. The same number of moles of NH3 is needed to produce NO2, as indicated by the equation 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g).
Therefore, Mass of NH3 = Moles of NH3 x Molar mass of NH3
= 2.38095x10^7 mol x 17 g/mol
= 4.0476x10^8 g
= 4.0476x10^5 kg
So, the correct mass of NH3 needed to produce 1.0x10^6 kg of HNO3 is 4.0476x10^5 kg.