Predict the moles of ammonia that can be made using 22.4 g of nitrogen and unlimited amount of hydrogen
This question refers to the chemical reaction for the synthesis of ammonia.
If you were not familiar with the reaction, you would start by finding the formulas. You can tell from the question that hydrogen and nitrogen are the reactants, and ammonia is the product. It is important to remember the diatomic gases. Hydrogen and nitrogen are both diatomic gases, as are all elements that are gases in their reference states, excluding noble gases. If you are not familiar with ammonia, you might have to look it up, and you would find the formula NH₃.
Thus far we have
`H_2 + N_2 -> NH_3`
The reaction needs to be balanced. The number of each kind of atom on the left side of the reaction arrow must be the same as the number of that kind of atom on the right. This is in keeping with Dalton's Theory: atoms are not created, destroyed, or changed in a chemical reaction. They are merely rearranged, so that all of the same atoms present before the reaction are also present after. In balancing, we only change the coefficients, representing the numbers of each chemical element or compound, NOT the subscripts, which would mean changing the identities of the chemicals present. We get
`3 H_2 + N_2 -> 2 NH_3`
This reaction actually goes to equilibrium, but the question is phrased as a stoichiometry problem, so we will treat the reaction as going to completion. We are to find the amount of ammonia, in moles, that can be made from a given amount, 22.4 grams, of nitrogen, in the presence of an excess of hydrogen.
We know that atoms and molecules combine in fixed ratios, producing amounts of products in fixed ratios to the reactants and to each other, and all of these ratios, or relative amounts, are represented by the coefficients in the balanced chemical equation. Because an unlimited amount of hydrogen is present, we don't have to worry about hydrogen. Nitrogen will react with the needed amount of hydrogen until the nitrogen is all used up, at which point no more ammonia can be made. The balanced equation shows us that when one molecule of nitrogen (`N_2`) reacts, two molecules of ammonia are produced. If two molecules of nitrogen react, four molecules of ammonia will be produced; three molecules of nitrogen will produce six molecules of ammonia, always in the same ratio of 1:2.
Since moles are a stand-in for molecules, the same relationship holds true for moles. We are given the amount of nitrogen in grams, however, so we need to convert it to moles. Since a nitrogen molecule contains two nitrogen atoms, its molar mass is twice the molar mass of atomic nitrogen:
molar mass of `N_2 = 2xx14.0067 g = 28.0134 g`
Our reaction starts with 22.4 g of nitrogen. We can convert to moles by multiplying this by the conversion factor (1 mole/28.013 g):
moles `N_2 = 22.4 g xx((1 mol)/(28.0134 g)) = 0.7996 mol`
where we have rounded to one extra digit, as is usual for intermediate steps in calculations. (This answer has three significant figures because the given quantity of 22.4 g has three significant figures.)
You may be thinking to yourself, "if one mole of nitrogen produces two moles of ammonia, then this number of moles of nitrogen produces twice this many moles of ammonia." If so, you are correct, and you may be developing a feel for how ratios and proportions work in chemistry.
If that is not clear, don't despair. There is a procedure to break this down into smaller, easier steps. We will create a table showing the quantity of each substance (hydrogen, nitrogen, and ammonia) at three stages: Before the reaction, the Change that takes place, and After the reaction. We are creating what is called a Before/Change/After table, or BCA table.
We need the balanced chemical equation close by so we can refer to it:
`3 H_2 + N_2 -> 2 NH_3` .
We list the chemical species at the top of the table. Below them, we label three lines B (before), C (change), and A (after). The Before line is for the initial conditions we are told about in the problem. Before the reaction, we know we have excess H₂, we have 0.7996 mol N₂, and no NH₃.
As the reaction has not started yet:
..........H₂........ N₂ ...........NH₃.
B excess ...0.7996 mol ...0 mol
Next we look at the Change (C) line. The basis of this problem is assuming that all of the nitrogen is used up in producing ammonia, so the change in amount of N₂ is -0.7996 mol (we can also fill in the bottom line, After, for N₂, showing 0 mol: it is all used up). We use a negative number for the change as it represents a decrease, an amount being subtracted.
B excess......0.7996 mole....0 mol
C ..............-0.7996 mol
A ...............0 mol
Here's where it gets interesting. The Before line represented the conditions set up in the lab and described in the problem, but the Change line represents what happens in the reaction. The Change line is where those fixed ratios expressed in the balanced chemical reaction come into play. In words, every time a given amount of N₂ is used up, three times that amount of H₂ is used up, and two times that amount of NH₃ is produced. This is based on the coefficients in the balanced chemical equation, which are always the same no matter what we do in the lab.
Now our table has
B excess ....................0.7996 mole ....0 mol
C -3*(0.7996 mol) ...-0.7996 mol .....+2*(0.7996 mol)=1.599 mol
A ...............................0 mol
We can fill in the final, After, line by adding and subtracting as appropriate. Note that an excess, or unlimited amount of hydrogen, minus some finite amount, is still excess. The amount of ammonia after the reaction is 0 mol + 1.599 mol = 1.599 mol, which we round to 1.60 mol, because it is our final answer and we know it has three significant figures. Thus, we get
B excess ...................0.7996 mol..... 0 mol
C -3*(0.7996 mol).... -0.7996 mol ...+2*(0.7996 mol)=1.599 mol
A excess ...................0 mol .............1.60 mol
Strictly speaking, the question asks for the amount of ammonia made, which is given on the Change line and would be the same even if we'd started with some ammonia already present. Notice that the answer is the same in both lines, the only difference being rounding. 1.60 Moles of ammonia can be made.
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