Consider the following generic chemical equation: 2W + 3X ? 3Y + Z
When 5 units of W and 6 units of X are allowed to react, the limiting reactant would be:
a)W
b)x
c)Y
d)Z
There are two ways to do this, the long way (which I prefer) and the short way (which most other people prefer).
The long way first:
2W + 3X ==> 3Y + Z
We are given 5W and 6X. Take these one at a time.
Use the coefficients in the balanced equation to convert units of W to units of either product. Let's choose Z.
5W x (1 unit Z/2 units W) = 5x(1/2) = 2.5 units Z.
Next we convert X.
6X x (1 unit Z/3 units X) = 6 x (1/2) = 2 units Z.
Both answers can't be right; the correct one is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, X is the limiting reagent and 2 units of Z will be formed.
Second method (and shorter):
Take EITHER W or X and convert to the other one. I'll do both so you can see how it works.
5W x (3 units X/2 units W = 5 x (3/2) = 15/2 = 7.5 units X; i.e., 5 units of W would require 7.5 units of X. Do we have that much? No, we have only 6; therefore, X must the limiting reagent.
Just to make sure we didn't goof, convert 6 units X to units of W.
6X x (2 units W/3 units X) = 6 x (2/3 = 4 units W. Do we have that much? Yes, therefore, the OTHER one (X) must the limiting reagent.
This may be more than you ever wanted t know about limiting reagent work BUT it's complete. Use either method you feel comfortable with.
There is a typo that confused me! "6X x (1 unit Z/3 units X) = 6 x (1/2) = 2 units Z. "
Well, in this chemical equation, it seems like we have plenty of options for reagents. But, it's a classic case of "too many cooks in the lab" -- I mean, "too many reactants in the beaker."
To find the limiting reactant, we need to compare the number of units of W and X available. You mentioned we have 5 units of W and 6 units of X.
Now, let's do some math. According to the equation, it takes 2 units of W and 3 units of X to make 3 units of Y and 1 unit of Z.
So, let's calculate how many units of Y can be formed from the available reactants:
- From W, we can make 5/2 units of Y
- From X, we can make 6/3 = 2 units of Y
Okay, so which one is smaller? 5/2 or 2?
Surprisingly, but not-so-surprisingly, 5/2 is bigger than 2! So, it means that our good old W is the limiting reactant here. The party's over for W, my friend!
Therefore, the answer is (a) W.
To determine the limiting reactant, we need to compare the number of moles of each reactant available with the stoichiometric coefficients in the balanced chemical equation.
For W: 5 units of W
For X: 6 units of X
To compare them, we need to convert the number of units into moles. Let's assume that the molar mass of W is MW and the molar mass of X is MX.
Number of moles of W = (5 units of W) / (MW g/mole)
Number of moles of X = (6 units of X) / (MX g/mole)
Let's simplify the equation by dividing both sides by the smallest coefficient:
2W + 3X -> 3Y + Z
Now, let's compare the moles of each reactant to their stoichiometric coefficients:
(W moles) / 2 = (X moles) / 3
If the ratio turns out to be less than 1, then W is the limiting reactant.
If the ratio turns out to be greater than 1, then X is the limiting reactant.
Calculate the mole ratio (W moles / X moles) and determine the limiting reactant.
To determine the limiting reactant in a chemical reaction, you must compare the number of moles (or units) of each reactant to the stoichiometric coefficients in the balanced chemical equation.
In this case, the balanced equation is:
2W + 3X → 3Y + Z
Given: 5 units of W and 6 units of X
To find the limiting reactant, we need to convert the units of W and X to moles. Once we have the moles of each reactant, we can compare them to the stoichiometric coefficients.
The molar ratio between W and X is 2:3. So, we need to determine which reactant (W or X) has the smallest number of moles when we convert their given units to moles.
Let's calculate the number of moles for W and X:
Moles of W = Given units of W / Stoichiometric coefficient of W
= 5 units / 2
= 2.5 moles
Moles of X = Given units of X / Stoichiometric coefficient of X
= 6 units / 3
= 2 moles
Comparing the number of moles, we find that there are 2.5 moles of W and 2 moles of X. Since X has a smaller number of moles compared to W, X is the limiting reactant.
Therefore, the answer is: b) X