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klasskru [66]
3 years ago
5

Nitric acid is produced commercially by the Ostwald process, represented by the following 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) What mass in kg of NH3 must be used to produce 7.6 ✕ 106 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?
Chemistry
1 answer:
Marrrta [24]3 years ago
4 0

Answer:

The answer to your question is: 1538095.2 kg of NH3

Explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

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

                            3(46) kg--------------   2(63) kg  

                                  x     ---------------  7600000 kg

           x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30            

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

                       2(30) ------------------2(46)

                        x        ---------------- 8323809.5 kg  

             x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

                      4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

                       4(17) -------------------- 4(30)

                         x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3

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Answer: 11.5 moles of carbon

Explanation:

Based on Avogadro's law:

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To get the value of Z, cross multiply:

(6.93 x 10^24 atoms x 1mole) = (6.02 x 10^23 atoms x Z moles)

6.93 x 10^24 = (6.02 x 10^23 x Z)

Z = (6.93 x 10^24) ➗ (6.02 x 10^23)

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