50g/28g= 1.7857 moles
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
The number of atoms is 45,166.10e23.
0,044 moles of NH3 can be produced.
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
Only one mole
I assume you mean this reaction. N2 + 3H2 --> 2NH3 0.90 moles N2 (3 moles H2/1 mole N2) = 2.7 moles hydrogen gas needed =====================
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
50 g of NaCl represent 0,856 moles.
The answer is 0,0043 moles of N2.
N2 + 3H2 -> 2NH3 If you have moles produced you can do it this way. 22.5 moles NH3 (3 moles H2/2 moles NH3) = 33.8 moles H2 needed -----------------------------------
2 x 0.60 = 1.2 the reaction is N2 + 3H2 -> 2NH3 (1 mole of nitrogen N2 give 2 moles of NH3)
The number of atoms is 45,166.10e23.
0,044 moles of NH3 can be produced.
The reaction of nitrogen with hydrogen to form ammonia is: N2 +3H2 = 2NH3 Therefore to make 10 moles of ammonia you need 5 moles N2 and 15 moles H2
In the equation N2+3H2=2NH3, the amount of ammonia produced from 50g of N would be 16.667g.
N2 + 3H2 --> 2NH3 You have been told, indirectly, that nitrogen limits and will drive the reaction. 3 moles N2 (2 moles NH3/1 mole N2) = 6 moles ammonia gas produced ========================
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3