Given the equation N2(g)+3H2(g)→2NH3(g), what is the total number of moles of NH3 produced when 10 mol of H2 reacts completely with N2?
1. 2.0 mol
2. 3.0 mol
3. 6.7 mol
4. 15 mol
6.7
6.7
This is done just like the Al/H2SO4 question.
10 mols H2 x (2 mols NH3/3 mol H2) = 10 x 2/3 = ?
3 mol
To determine the total number of moles of NH3 produced when 10 mol of H2 reacts completely with N2, we need to use the balanced equation and the mole ratio between the reactants and products.
The balanced equation is:
N2(g) + 3H2(g) → 2NH3(g)
From the equation, we can see that for every 3 moles of H2, 2 moles of NH3 are produced. This gives us a mole ratio of 2:3 between NH3 and H2.
Since we have 10 mol of H2, we can set up a proportion using the mole ratio:
(2 mol NH3 / 3 mol H2) = (x mol NH3 / 10 mol H2)
By cross-multiplying and solving for x, we can find the number of moles of NH3 produced:
(x mol NH3) = (2 mol NH3 / 3 mol H2) * 10 mol H2
x = 20/3 = 6.7 mol
Therefore, the total number of moles of NH3 produced when 10 mol of H2 reacts completely with N2 is approximately 6.7 mol.
So, the answer is option 3. 6.7 mol.