50g/28g= 1.7857 moles
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
0,044 moles of NH3 can be produced.
The number of atoms is 45,166.10e23.
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
In the equation N2+3H2=2NH3, the amount of ammonia produced from 50g of N would be 16.667g.
To determine the number of moles in 50g of NaCl, you first need to find the molar mass of NaCl, which is approximately 58.44 g/mol. Then, you divide the given mass by the molar mass to find the number of moles. So, 50g of NaCl represents approximately 0.855 moles (50g / 58.44 g/mol = 0.855 mol).
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
The answer is 0,0043 moles of N2.
Assuming a balanced chemical equation, you would need 3 moles of H2 to react with 1 mole of N2. Therefore, if you have 0.90 moles of N2, you would need 0.90 x 3 = 2.70 moles of H2 to fully react with it.
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 -----------------------------------
0,044 moles of NH3 can be produced.
The number of atoms is 45,166.10e23.
2 x 0.60 = 1.2 the reaction is N2 + 3H2 -> 2NH3 (1 mole of nitrogen N2 give 2 moles of NH3)
0,3 moles of nitrogen reacted.
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