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. -----
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?
1 answer
1 answer