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To find the grams of F2 required for the reaction, first calculate moles of NH3: 69.3 g NH3 / 17.03 g/mol = 4.07 moles NH3. From the balanced equation, 5 moles of NH3 react with 2 moles of F2, so you need 4.07 moles NH3 * (2 moles F2 / 5 moles NH3) * 38.0 g/mol = 30.6 g of F2 for complete reaction.
The balanced chemical equation for the reaction between F2 and NH3 is: 3 F2 + 4 NH3 -> 6 HF + N2 From the equation, we can see that 3 moles of F2 react with 4 moles of NH3. To find the moles of F2 required to react with 3.50 moles of NH3, we can set up a proportion: 3 moles F2 / 4 moles NH3 = x moles F2 / 3.50 moles NH3 Solving for x, we find that 2.625 moles of F2 are required. To convert this to grams, we use the molar mass of F2 which is approximately 38.00 g/mol. 2.625 moles F2 x 38.00 g/mol = 99.75 grams of F2 required to react with 3.50 moles of NH3.
The bond energy of NH3, which is the energy required to break one mole of NH3 molecules into its individual atoms, is approximately 391 kJ/mol.
Atomic mass of NH3 is 17.So there are 1.353mol.
You need the balanced chemical equation for N2 and H2 combining to form ammonia, NH3.N2 (g) + 3 H2 (g) -----> 2 NH3 (g)Moles NH3 = ( 55.5 g NH3 ) / ( 17.03 g/mol NH3 ) = 3.259 moles of NH3n N2 required = ( 3.259 mol NH3 ) ( 1 N2 mol / 2 NH3 mol ) = 1.629 moles N2m N2 required = ( 1.629 mol N2 ) ( 28.103 g N2 / mol N2 ) = 45.67 g N2 needed
Only 1 proton has NH3
There is one nitrogen atom in the molecule NH3.
There is not element that contains NH3, NH3 is a compound made of two different elements. An elements is made of nothing but itself.
0,044 moles of NH3 can be produced.
There is a total of 1 nitrogen atom in a molecule of NH3.
4 atoms
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