The subject is stoichiometry. How many moles of NH3 can be produced from 12.0mol of H2 and excess N2?
First, we write the balanced chemical equation involved. Since the reaction is synthesis of NH3,
N2 + 3 H2 -> 2 NH3
Since N2 is excess, H2 is limiting and we therefore use its moles to determine the amount of NH3 produced. In the equation, there are 2 moles of NH3 produced over 3 moles of H2 reactant. Thus, we use this ratio:
12.0 mol H2 ( 2 mol NH3 / 3 mol H2 ) = 8.00 mol NH3
Hope this helps~ :3
Jai ... what happen to the excess N2
To determine the number of moles of NH3 that can be produced from 12.0 moles of H2 and excess N2, we need to use the balanced chemical equation for the reaction between H2 and N2 to form NH3:
3H2 + N2 -> 2NH3
From the balanced equation, we see that it takes 3 moles of H2 to produce 2 moles of NH3. This means that the stoichiometric ratio between H2 and NH3 is 3:2.
Given that we have 12.0 moles of H2, we can set up a proportion to calculate the moles of NH3:
(12.0 moles H2) / (3 moles H2) = (x moles NH3) / (2 moles NH3)
Simplifying the proportion:
4 = (x moles NH3) / 2
Now we can solve for x, which represents the moles of NH3:
4 * 2 = x
8 = x
Therefore, 8 moles of NH3 can be produced from 12.0 moles of H2 and excess N2.
To calculate the moles of NH3 that can be produced in a stoichiometric reaction, we need to know the balanced chemical equation. In this case, the balanced chemical equation is:
N2 + 3H2 -> 2NH3
From the equation, we can see that 3 mol of H2 reacts with 1 mol of N2 to produce 2 mol of NH3. This means that the stoichiometric ratio of H2 to NH3 is 3:2.
Given that we have 12.0 mol of H2, we can set up a ratio to find the moles of NH3 produced:
(12.0 mol H2) x (2 mol NH3 / 3 mol H2) = 8.00 mol NH3
Therefore, 8.00 moles of NH3 can be produced from 12.0 moles of H2 and excess N2.