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will the reactant with less moles always be the LR? do the coefficients of the reactants matter?
limiting reagent
- Thread starter Oh_Gee
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The substrate that is used up first.
So it depend on coefficients.
For example, 10A + B -----> A10B
If there are 10 moles of A and 5 moles of B, the LR will be A even though it has more moles.
The reaction will use 10 moles of A and only 1 of B, leaving 4 moles of B.
So it depend on coefficients.
For example, 10A + B -----> A10B
If there are 10 moles of A and 5 moles of B, the LR will be A even though it has more moles.
The reaction will use 10 moles of A and only 1 of B, leaving 4 moles of B.
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so if you were given grams of A and B and you calculated that you have 1 mol A and 3 mols B, A would still be the LR? and .1 mols of A10B will be made?
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Yup exactly.
You can calculate your moles and then use dimensional analysis to check for the limiting reagent.
Given 1 mol A and 3 mol B
1 mol A * (1 mol A10B/10 mol A) = .1 mol A10B
3 mol B * (1 mol A10B/1 mol B) = 3 mol B
You can only go as far as your weakest link!
So then we can say that a reaction creating .1 mol A10B * (1 mol B/1 mol A10B) = .1 mol B will be used
So we started with 3 mol B and used .1 mol B = 2.9 mol B remaining.
You can calculate your moles and then use dimensional analysis to check for the limiting reagent.
Given 1 mol A and 3 mol B
1 mol A * (1 mol A10B/10 mol A) = .1 mol A10B
3 mol B * (1 mol A10B/1 mol B) = 3 mol B
You can only go as far as your weakest link!
So then we can say that a reaction creating .1 mol A10B * (1 mol B/1 mol A10B) = .1 mol B will be used
So we started with 3 mol B and used .1 mol B = 2.9 mol B remaining.
