Question

Consider the generic reaction between reactants A and B:
3A+4B→2C
If a reaction vessel initially contains 9 mol A and 4 mol B, how many moles of A will be in the reaction vessel once the reactants have reacted as much as possible? (Assume 100% actual yield.)
How many moles of B will be in the reaction vessel once the reactants have reacted as much as possible? (Assume 100% actual yield.)
How many moles of C will be in the reaction vessel once the reactants have reacted as much as possible? (Assume 100% actual yield.)
Express your answer using two significant figures.

Answer 1

The question makes no sense without information about how much A and B you started with.

Answer 2

I fixed it. Thank you for catching that

Answer 3

This is a good question to think about stoichiometry (amounts needed for reaction).
Equation: 3A + 4B -> 2C
What you have: 9 mol A, 4 mol B. Assuming 100% yield, the equation means for every 3 moles of A, you can react with exactly 4 moles of B to produce exactly 2 moles of C.

So let's think, if we have 9 moles of A, how much we'd use. This is pretty easy, multiply the equation by 3.

3A + 4B -> 2C (x 3) --> 9A + 12B -> 6C.

This means if we want to use 9 moles of A completely, we need 12 moles of B.
However, we are only given 4 moles of B meaning we don't have enough B to make 12 moles. So what happens is even though we have 9 moles of A, we can't use all of it because we are limited by how much of B we have. B is termed a 'limiting reagent' because it limits how much A we can use.

So how much A can we actually use then? Well go back to our original equation again.
3A + 4B -> 2C. So for 4 moles of B we can use 3 moles of A. That means we can use up all of B and have enough A to use, and have some A left over. With 9 moles of A, we only use 3 moles so we have 6 moles of A leftover, and this is termed 'excess'.

Finally how much C do we have? Well we used up 4 of B and 3 of A, we expect to get 2 of C. So in summary if we use 9A, 4B in the reaction, because the ratios of A and B are not in stoichiometric proportion (i.e. the number of moles fits the numbers on the original equation exactly), this means you will have some things leftover in excess and a limiting reagent. In this case, all of B could be used (so 0 mol left after reaction is complete) but we ended up with leftover A of 6 moles, and 2 moles of C according to the equation.

Answer 4

By the way, another helpful way to think of this is to do a ratio of moles of A to B.
Theoretical: A/B = 3/4
Actual A/B = 9/4
As actual A/B > theoretical A/B, this means we will have more of A compared to B finally and this will be the excess, while B will be the limiting reagent. We always use up all the limiting reagent, so the calculations for the moles of excess and the product depend on the amount of limiting reagent used.

Answer 5

Thank You so much, that makes more sense

Answer 6

This one's especially easy, since we're given the 4 moles of B that the reaction requires as written. That will use 3 moles of A; we have 9, so there will be 6 left. And we get 2 moles of C, per the equation.