Question

What mass of NH3 in grams must be used to produce 1.81 tons of HNO3 by the Ostwald process, assuming an 80.0 percent yield in each step (1 ton=2000 lb; 1 lb= 453.6 g)?

Answer 1

The three reactions involved in the Ostwald process for the conversion of NH3 to HNO3 are:

4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(l)   ------(1)

2NO(g) + O2(g) ==> 2NO2(g)  ------------------------(2)

3NO2(g) + H2O(l) ==> 2HNO3(aq) + NO(g) ---(3)

Mass of HNO3 produced = 1.81 tons

In grams, the mass of HNO3 = 1.81*2000*453.6 = 1642032 g

Molar mass of HNO3 = 63 g/mol

Thus, # moles of HNO3 produced = 1642032/63 = 26064 moles

Based on the stoichiometry of reaction (3):

Theoretical moles of NO2 = 1.5(Moles HNO2 ) = 1.5(26064) = 39096

Assuming 80% yield:

Actual moles of NO2 = 0.8*39096 = 31277 moles

Based on the stoichiometry in reaction (2):

Theoretical moles of NO = 31277 moles

Assuming 80% yield:

Actual moles of NO = 0.8*31277 = 25022 moles

Bases on stoichiometry in reaction(3):

Moles of NH3 = 25022 moles

Assuming 80% yield:

Actual moles of NH3 required = 0.8*25022 = 20018 moles

Mass of NH3 required = 20018 moles * 17 g/mole = 340306 g = 340.31 kg

Answer 2

Answer: Mass of $NH_3$ used is 1730812.5 grams.

Explanation: Ostwald's process is the process of the production of nitric acid from ammonia. Steps for the process are:

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(l)$

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

$2NO_2(g)+H_2O(l)\rightarrow HNO_3(aq.)+HNO_2(aq.)$

Total mass of $HNO_3$ produced = 1.81 tons

Using conversion factor, we get

Total mass of $HNO_3$ produced = (1.81 × 2000 × 453.6)g = 1642032 g

Moles of $HNO_3$ is calculated by using the formula:

$\text{Number of Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Molar mass of nitric acid = 63 g/mol

$\text{Number of Moles}=\frac{1642032g}{63g/mol}=26064moles$

We are given that the percentage yield is 80 % in every step, then

Moles of $NO_2$, theorectically (From Step-III) = 2 × 26064 = 52128 moles

Actual moles of $NO_2=52128\times \frac{100}{80}=65160moles$

Now, theoretical moles of NO (in Step-II) = Actual moles of $NO_2$

Theoretical moles of NO = 65160 moles

Actual moles of NO = $65160\times \frac{100}{80}=81450moles$

Now, theoretical moles of $NH_3$ (in Step-I) = Actual moles of NO

Theoretical moles of $NH_3$ = 81450 moles

Actual moles of $NH_3$ = $81450\times \frac{100}{80}=101812.5moles$

To calculate the mass of $NH_3$, we use equation 1, we get

Molar mass of $NH_3$ = 17 g/mol

$\text{Mass of }NH_3\text{ used}=101812.5mol\times 17g/mol=1730812.5g$