Question

Nitric acid is produced commercially by the Ostwald process,represented by the following equations:

4NH3(g) + 5O2(g) -----> 4NO(g) +6H2O
2NO(g) + O2(g) ------> 2NO2(g)
3NO2(g) + H2O(l) -------->2HNO3(aq) + NO(g)
What mass of NH3 must be used to produce 1.0 x106 kg HNO3 by the Ostwald process? Assume100% yield in each reaction and assume the NO produced in the thirdstep is not recycled.

Answer 1

To produce 1.0 x 10¹⁶ kg of HNO3 in the Ostwald process, 4.056 x 10⁵ kg of NH3 is required, following stoichiometric calculations based on balanced chemical reactions and assuming 100% yield.

Step-by-step explanation:

To calculate the mass of NH3 needed to produce 1.0 x 106 kg of HNO3 in the Ostwald process, we need to use the stoichiometry of the balanced chemical equations provided. From the final equation (c), we see that 3 moles of NO2 yield 2 moles of HNO3. The molar mass of NH3 (17.03 g/mol) and HNO3 (63.01 g/mol) are also needed for the calculation.

First, we convert the mass of HNO3 to moles:
1.0 x 106 kg HNO3 * (1 x 103 g/kg) * (1 mol/63.01 g) = 1.587 x 107 moles of HNO3.

Next, from the balanced equation (c), we use the stoichiometry to find moles of NH3:
1.587 x 107 moles HNO3 * (3 moles NO2/2 moles HNO3) * (4 moles NH3/4 moles NO) = 2.381 x 107 moles of NH3.

Finally, we convert the moles of NH3 to mass:
2.381 x 107 moles NH3 * 17.03 g/mol = 4.056 x 108 g of NH3.

Converting grams to kilograms, we get:
4.056 x 108 g * (1 kg/1 x 103 g) = 4.056 x 105 kg of NH3.

Therefore, 4.056 x 105 kg of NH3 is required to produce 1.0 x 106 kg of HNO3, assuming 100% yield in the Ostwald process.