Assume that you had 15 nuts (N) and 10 bolts (B). How many N2B molecules could you make by screwing two nuts onto each bolt?

There are two nuts for each bolt. We have 15 nuts so we can make 7.

I'm supposed to "Show all calculation for full credit." Is there anyway of setting up an algebraic equation to demonstrate the work I did?

You want to make N2B with two nuts on each bolt. Write a chemical equation.

2N + B = N2B (combo)
The equation is balanced.
Now try 10B and 15N to see how many N2B can be formed. Just like a stoichiometry problem.
15N x (1N2B/2N) = 7.5 and of course we can't have half a unit so that gives 7.0.
10B x (1N2B/B) = 10.
Therefore, 15 nuts is the limiting reagent; 7.0 is what can be formed.

You can show this in diagram fashion if you wish. Let + be nuts and ---- be the bolt.
1. +----+
2. +----+
3. +----+
4. +----+
5. +----+
6. +----+
7. +----+
7.0 full units.

8. half unit. +----
9. -----
10. -----

Yes, you can set up an algebraic equation to demonstrate the work you did. Let's denote the number of nuts as N and the number of bolts as B.

Given that there are 15 nuts (N) and 10 bolts (B), and each bolt requires two nuts, you can express the relationship between N and B as:

2N = B

To find the maximum number of N2B molecules that can be formed, we need to determine the maximum number of bolts (B) that can be formed with the given number of nuts (N).

Since there are 15 nuts, we can substitute N=15 into the equation:

2(15) = B

Simplifying the equation:

30 = B

Therefore, the maximum number of bolts (B) that can be formed is 30.

Now, to find the maximum number of N2B molecules, we can divide the number of bolts (B) by 2 (since each bolt requires 2 nuts):

30 / 2 = 15

Therefore, you can make 15 N2B molecules by screwing two nuts onto each bolt.

To summarize, the algebraic equation you can use to represent the problem is 2N = B, and by substituting N=15, you can find B=30. Then, by dividing B by 2, you can find the maximum number of N2B molecules, which in this case is 15.