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.]

Convert 1.0 x 10^6 kg HNO3 to mols HNO3. I assume you know mols = grams/molar mass.

Then convert mols HNO3 to mols NO2 using the coefficients in the last equation (equation 3).

You must have omitted the third step since that was to be ignored.

Convert mols NO2 to mols NO using the coeffieicnts in the second equation.

Convert mols NO to mols NH3 using the coefficients in the first equation.

Finally, convert mols NH3 to grams NH3 remembering that grams = mols x molar mass.

Post your work if you get stuck.

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?

74757578

this is correct?

To convert 1.0 x 10^6 kg HNO3 to moles HNO3, we can use the formula:

moles = (mass / molar mass)

The molar mass of HNO3 is calculated as follows:
H = 1 atomic mass unit (amu)
N = 14 amu
O = 16 amu (3 oxygen atoms in HNO3)

Molar mass of HNO3 = (1 amu + 14 amu + (16 amu * 3)) = 63 amu

Now we can calculate the moles of HNO3:
moles = (1.0 x 10^6 kg / 63 amu) = 15873.01587 mol HNO3

Next, we'll use the stoichiometry of the reactions to calculate the moles of NH3 needed:

According to the balanced equations:
4 NH3(g) + 5 O2(g) > 4 NO(g) + 6 H2O(g)
2 NO(g) + O2(g) > 2 NO2(g)
3 NO2(g) + H2O(l) > 2 HNO3(aq) + NO(g)

From the second equation, 2 moles of NO2 are produced from 1 mole of NO.
From the first equation, 4 moles of NO are produced from 4 moles of NH3.

Therefore, 4 moles of NH3 are needed to produce 2 moles of NO2.
And as we calculated before, 15873.01587 moles of HNO3 are produced.

To find the moles of NH3 needed, we use the molar ratios:
(4 mol NH3 / 4 mol NO) * (2 mol NO / 15873.01587 mol HNO3) = 0.00025123864 mol NH3

Finally, we can calculate the mass of NH3 needed using the molar mass of NH3:
mass = (0.00025123864 mol NH3) * (17 amu / 1000) = 0.00427624588 kg NH3

Therefore, approximately 0.0043 kg of NH3 must be used to produce 1.0 x 10^6 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction

Let's go through the calculations step by step to confirm if the answer is correct.

1. Convert 1.0 x 10^6 kg HNO3 to moles HNO3.
Since mols = grams / molar mass, we can find the moles of HNO3 using its molar mass. The molar mass of HNO3 is: 1(1) + 1(14) + 3(16) = 63 g/mol. Therefore, the moles of HNO3 is:
1.0 x 10^6 kg / 63 g/mol = 15873015.87 mol.

So far, your calculation is correct.

2. Convert moles HNO3 to moles NO2 using the coefficients in the equation 3.
From the equation 3, we see that it takes 3 moles of NO2 to produce 2 moles of HNO3. Therefore, we can set up the following proportion:
3 mol NO2 / 2 mol HNO3 = x mol NO2 / 15873015.87 mol HNO3.

Solving for x, we get:
x = (3 mol NO2 / 2 mol HNO3) * 15873015.87 mol HNO3 = 23809523.81 mol NO2.

So the moles of NO2 is indeed 23809523.81 mol.

3. Convert moles NO2 to moles NO using the coefficients in the equation 2.
From the equation 2, we see that it takes 2 moles of NO to produce 2 moles of NO2. Therefore, the moles of NO is equal to the moles of NO2.

So the moles of NO is also 23809523.81 mol.

4. Convert moles NO to moles NH3 using the coefficients in the equation 1.
From the equation 1, we see that it takes 4 moles of NH3 to produce 4 moles of NO. Therefore, the moles of NH3 is also equal to the moles of NO.

So the moles of NH3 is 23809523.81 mol.

5. Convert moles NH3 to grams NH3 using the molar mass.
The molar mass of NH3 is: 1(14) + 3(1) = 17 g/mol. Therefore, the mass of NH3 is:
23809523.81 mol NH3 * 17 g/mol = 404761904.8 g.

Converting grams to kilograms:
404761904.8 g / 1000 = 404761.904 kg.

Therefore, the correct answer is 404761.904 kg of NH3.

It seems that there was an error in your calculation. Please double-check your steps and calculations to find the mistake.