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| 1. |
{ The mass of }N_2F_4 produced by the reaction of }}{1 mole of }NH_3 and }2.5 moles of }F_2 is }35.6g . Hence, }} the percentage yield of the given reaction is }{2NH_3+5F_2→ N_2F_4+6HF |
| Answer» { The mass of }N_2F_4 produced by the reaction of }}{1 mole of }NH_3 and }2.5 moles of }F_2 is }35.6g . Hence, }} the percentage yield of the given reaction is }{2NH_3+5F_2→ N_2F_4+6HF | |