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?

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

beta decay is the answer