Ammonia gas is produced commercially from the reaction ofnitrogen and hydrogen.

Question

Ammonia gas is produced commercially from the reaction ofnitrogen and hydrogen. What volume of ammonia can be produced fromthe reaction of 5.5 x 103 kg of N2 and 1.5 x103 kg of H2? Assume the reaction is100% efficient and the product is collected at 325k and25atm. N2(g) + 3H2(g) 2NH3(g) Ammonia gas is produced commercially from the reaction ofnitrogen and hydrogen. What volume of ammonia can be produced fromthe reaction of 5.5 x 103 kg of N2 and 1.5 x103 kg of H2? Assume the reaction is100% efficient and the product is collected at 325k and25atm. N2(g) + 3H2(g) 2NH3(g)

Explanation / Answer

Number of moles of N2 present in the mixture= weight of the N2 / M.Wt of N2                                                                       =5.5*106 g / 28g/mole                                                                      = 1.964*105 mole Number of moles of H2 present   =1.5 x 106 / 2g/mole                                                    = 7.5*105 mole Threfore according to given reaction 1 mole of the N2reacted with the 3moles of H2.         Theefore moleration of the N2 and H2 in the present raction                                       = 7.5*105 mole H2 /1.964*105 mole                                       = 3.81 mole H2 / 1 mole N2 Therefore number of moles of ammonia produced =2*1.964*105 mole                                                                             = 3.928*105 mole The volume of the ammonia produced = nRT/P                                                             =3.928*105 mole * 0.0821Latm mol-1K-1 * 325K /25atm                                                            = 4.19*105 L                                                                       =5.5*106 g / 28g/mole                                                                      = 1.964*105 mole Number of moles of H2 present   =1.5 x 106 / 2g/mole                                                    = 7.5*105 mole Threfore according to given reaction 1 mole of the N2reacted with the 3moles of H2.         Theefore moleration of the N2 and H2 in the present raction                                       = 7.5*105 mole H2 /1.964*105 mole                                       = 3.81 mole H2 / 1 mole N2 Therefore number of moles of ammonia produced =2*1.964*105 mole                                                                             = 3.928*105 mole The volume of the ammonia produced = nRT/P                                                             =3.928*105 mole * 0.0821Latm mol-1K-1 * 325K /25atm                                                            = 4.19*105 L

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