Question

use graph paper or graphing software to plot the moles of ammonia produced versus moles of each reactant used. Show each reactant as its own line.

Answer 1

Start from the balanced equation and the mole ratios:
N2 + 3 H2 → 2 NH3
So
- 1 mol N2 can produce 2 mol NH3 → NH3 = 2·(moles N2)
- 3 mol H2 produce 2 mol NH3 → NH3 = (2/3)·(moles H2)

Equations to plot (two separate lines on the same axes)
- y = 2x (x = moles N2, y = moles NH3)
- y = (2/3)x (x = moles H2, y = moles NH3)

Example data points (choose x from 0 to 5 mol):

Moles N2 used (x) -> NH3 produced (y = 2x)
0 → 0
0.5 → 1.0
1 → 2.0
2 → 4.0
3 → 6.0
4 → 8.0
5 → 10.0

Moles H2 used (x) -> NH3 produced (y = (2/3)x)
0 → 0
0.5 → 0.333...
1 → 0.666...
2 → 1.333...
3 → 2.0
4 → 2.666...
5 → 3.333...

How to plot
- On graph paper: draw x-axis (moles of reactant) and y-axis (moles NH3). Choose same x-scale for both reactants or make two separate x-labels if you prefer. Plot the N2 points and connect them with a straight line (slope 2). Plot the H2 points and connect (slope 2/3). Label each line and the axes, and mark the slope or equation on the graph.
- In Excel / Google Sheets: put x values in column A, compute column B = 2*A, column C = (2/3)*A. Insert a scatter plot with straight lines connecting points; include both columns B and C as series and label them.
- In Desmos or other graphing software: enter y = 2x and y = (2/3)x; set a reasonable x-range (0 to 5 or 0 to 10).

Note on limiting reagent
If you have both reactants present in arbitrary amounts, the actual NH3 produced equals the smaller of the two values given by the formulas above:
NH3 produced = min(2·n(N2), (2/3)·n(H2))
Graphically this is the lower of the two lines at a given pair of reactant amounts.