How many grams of ammonia are produced when 6.50 mol hydrogen reacts with nitrogen?
3H2 + N2 2NH3
2 mol NH3 for every 3 molsH2
6.5 (2/ 3) = 4.33 mols N2
mol NH3 = 14+3=17 grams per mol
17 * 4.33
To find out how many grams of ammonia (NH3) are produced when 6.50 mol of hydrogen (H2) reacts with nitrogen (N2), we need to use the given balanced chemical equation and the molar ratio between H2 and NH3.
The balanced chemical equation is:
3H2 + N2 -> 2NH3
From the equation, we can see that for every 3 moles of H2, we get 2 moles of NH3. Therefore, the molar ratio between H2 and NH3 is 3:2.
First, we need to determine the number of moles of NH3 produced. Since we have 6.50 mol of H2, we can set up a ratio based on the molar ratio:
(6.50 mol H2) x (2 mol NH3 / 3 mol H2) = 4.33 mol NH3
Now we know that 6.50 mol of H2 will produce 4.33 mol of NH3.
Next, we can calculate the mass of NH3 produced using the molar mass of NH3, which is the sum of the atomic masses of one nitrogen (N) atom and three hydrogen (H) atoms:
Molar mass NH3 = (1 mol N x atomic mass N) + (3 mol H x atomic mass H)
The atomic masses are:
Atomic mass N = 14.01 g/mol
Atomic mass H = 1.01 g/mol
Molar mass NH3 = (1 mol x 14.01 g/mol) + (3 mol x 1.01 g/mol) = 17.03 g/mol
Finally, we can calculate the mass of NH3 produced:
Mass NH3 = (4.33 mol NH3) x (17.03 g/mol NH3) = 73.6 g
Therefore, when 6.50 mol of H2 reacts with N2, approximately 73.6 grams of NH3 are produced.