A chemical engineer calculated that 15.0 mol H2 was needed to react with excess N2 to prepare 10.0 mol NH3. But the actual yield is 60.0% . Write a balanced chemical equation for the reaction. Is the amount of H2 needed to make 10.0 mol NH3 more than, the same as, or less than 15 mol? How many moles of H2 are needed?
15moles exactly
The balanced chemical equation for the reaction of hydrogen gas (H2) and nitrogen gas (N2) to produce ammonia gas (NH3) is as follows:
3H2 + N2 → 2NH3
To determine whether the amount of H2 needed to produce 10.0 mol of NH3 is more than, the same as, or less than 15.0 mol, we need to compare the stoichiometric ratio of H2 to NH3 in the balanced chemical equation.
From the balanced equation, we can see that 3 moles of H2 react to form 2 moles of NH3.
Therefore, the molar ratio between H2 and NH3 is:
3 mol H2 : 2 mol NH3
To find out how many moles of H2 are needed to produce 10.0 mol of NH3, we can set up a proportion:
3 mol H2 / 2 mol NH3 = x mol H2 / 10.0 mol NH3
Simplifying the proportion:
3/2 = x/10.0
Cross-multiplying:
2x = 3 * 10.0
2x = 30
Dividing both sides by 2:
x = 15
Therefore, 15.0 mol of H2 are needed to produce 10.0 mol of NH3.
Since the actual yield of the reaction is 60%, the amount of H2 that would actually react to produce 10.0 mol of NH3 is:
0.60 * 15.0 mol = 9.0 mol
So, the actual amount of H2 needed to produce 10.0 mol of NH3 is 9.0 mol.
To write a balanced chemical equation for the reaction, we need to know the reactants and products involved. In the given information, it is mentioned that hydrogen gas (H2) reacts with nitrogen gas (N2) to produce ammonia (NH3). Therefore, the balanced chemical equation for this reaction is:
3H2 + N2 -> 2NH3
Now, let's determine whether the amount of H2 needed to make 10.0 mol of NH3 is more than, the same as, or less than 15 mol.
According to the stoichiometry of the reaction, 3 moles of hydrogen gas (H2) are required to produce 2 moles of ammonia (NH3).
Given:
Actual yield = 60.0%
From the given information, the chemical engineer calculated that 15.0 mol of H2 was needed. However, the actual yield is only 60.0%. Therefore, the actual amount of NH3 produced would be:
Actual yield of NH3 = 60.0% of 10.0 mol = 0.60 * 10.0 mol = 6.0 mol NH3
Now, using the mole ratio in the balanced equation, we can calculate the required amount of H2:
(3 mol H2 / 2 mol NH3) * (6.0 mol NH3) = 9.0 mol H2
So, the amount of H2 needed to produce 10.0 mol of NH3 is 9.0 mol.
To summarize, the amount of H2 needed to make 10.0 mol of NH3 is less than the initial calculation of 15.0 mol. The actual amount required is 9.0 mol of H2.