What is the maximum mass, in grams, of NH3 that can be produced by the reaction of 1.0 g of N2 and 1.0 g of H2 using the reaction below?
N2+H2→NH3 (not balanced)
A) 0.60 g
B) 5.7 g
C) 1.2 g
D) 2.9 g
DrBob222 Please let me know if this is correct
N2 + 3 H2 → 2 NH3
(1.0 g N2) / (28.01344 g N2/mol) = 0.03570 mol N2
(1.0 g H2) / (2.01588 g H2/mol) = 0.4961 mol H2
0.03570 mole of N2 would react completely with 0.03570 x (3/1) = 0.1071 mole of H2, but there is more H2 present than that, so H2 is in excess and N2 is the limiting reactant.
(0.03570 mol N2) x (2 mol NH3 / 1 mol N2) x (17.03056 g NH3/mol) = 1.2 g NH3
That's my answer also.
Well, let's see if we can figure this out while keeping the mood light and breezy.
First things first, we need to balance the equation to get a better understanding of the stoichiometry. So, let me grab my balancing act skills and balance the equation for you.
N2 + 3H2 -> 2NH3
Now that we have a balanced equation, we can calculate the maximum mass of NH3 produced by using the concept of limiting reactants.
To do that, we need to determine which reactant limits the amount of product formed.
We can start by converting the masses of N2 and H2 into moles.
Using the molar masses of N2 (28 g/mol) and H2 (2 g/mol), we find that 1.0 g of N2 is equal to approximately 0.036 mol, and 1.0 g of H2 is equal to approximately 0.5 mol.
Since the stoichiometry of the reaction is 1:3 for N2 and H2 respectively, it means that for every 1 mol of N2, we need 3 mol of H2.
But we have 0.036 mol of N2, which means we would need (3 x 0.036) = 0.108 mol of H2 to react completely.
But we only have 0.5 mol of H2, which is more than enough to react with the available amount of N2.
So, H2 is in excess, and N2 is the limiting reactant.
Now we can calculate the maximum mass of NH3 produced by using the molar mass of NH3 (17 g/mol).
0.036 mol of N2 will produce (2 x 0.036) = 0.072 mol of NH3.
So, the maximum mass of NH3 produced is (0.072 mol x 17 g/mol) = 1.224 g.
Well, well, well, it looks like the correct answer is C) 1.2 g. Isn't chemistry just full of surprises? Keep in mind that this is the maximum and assumes a complete reaction, without any hiccups or spilled potions.
To solve this question, we need to determine the limiting reactant and then calculate the maximum mass of NH3 that can be produced based on the stoichiometry of the balanced equation.
1. Start by balancing the equation:
N2 + 3H2 → 2NH3
2. Calculate the molar mass of each substance:
N2 = 28.02 g/mol
H2 = 2.02 g/mol
NH3 = 17.03 g/mol
3. Convert the given masses of N2 and H2 to moles:
N2: 1.0 g / 28.02 g/mol = 0.0357 mol
H2: 1.0 g / 2.02 g/mol = 0.4950 mol
4. Now, we need to determine the limiting reactant. The limiting reactant is the one that will be completely consumed, therefore determining the maximum amount of product that can be formed. To find the limiting reactant, we compare the mole ratios of N2 and H2 from the balanced equation:
Ratio of N2 to NH3 = 1:2
Ratio of H2 to NH3 = 3:2
Based on the ratios, we see that 1 mole of N2 can produce 2 moles of NH3, while 3 moles of H2 can produce only 2 moles of NH3. Therefore, the limiting reactant is N2.
5. Calculate the theoretical yield of NH3 using the mole ratio between N2 and NH3:
0.0357 mol N2 x (2 mol NH3 / 1 mol N2) = 0.0714 mol NH3
6. Convert the moles of NH3 to grams using the molar mass of NH3:
0.0714 mol NH3 x 17.03 g/mol = 1.21 g
Therefore, the maximum mass of NH3 that can be produced is 1.21 g.
The correct answer is C) 1.2 g.
the balanced equation is a 2 in front of the N2
I worked this for someone just a day or so ago but I can't find it. Here is a summary.
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
N2 + 3H2 ==> 2NH3
mols N2 = grams/molar mass = ?
mols H2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols N2 to mols NH3.
Do the same and convert mols H2 to mols NH3.
You can see that mols NH3 produced is not the same; one of the answers must be wrong. The correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR.
Now use the LR value and convert that mols to grams.
g NH3 = mols NH3 x molar mass NH3