Question

How to find the limiting reagent in any problem. Anybody know any easy steps to understand?

Answer 1

The limiting reagent is the one that is completely consumed. I can better help you if you give me an example.

Answer 2

Yeah I get that, but I just totally forgot how to work it out. Let me find an example..

Answer 3

Ok..
22g of KOH + 30g of HBr.. What mass of KBr will be produced and what is the limiting reagent?

Answer 4

There are a couple of ways to solve this. I'll walk you through what I find the easiest.

First, let's write a stoichiometric reaction. \[KOH + HBr = KBr + H_2O\]
Second, find the molecular mass of each reaction species and each product species. \[\left[\begin{matrix}KOH & 39+1+16 = 56 \\ HBr& 1 + 80 = 81 \\ KBr & 39 + 80 = 119 \\ H_2O & 1 + 16+ 16 = 33\end{matrix}\right]\]
Third, determine the number of moles of the reaction species from molecular weight and given mass. \[N = {m \over MM}\]\[\left[\begin{matrix}KOH & 22/56 = 0.3929 \\ HBr & 30/81 = 0.3704\end{matrix}\right]\]
Fourth, from stoichiometric equation determine the stoichiometric ratio of KOH to HBr. This ratio is 1:1 because 1 mole of each reacts.
Fifth, determine the actual ratio of KOH to HBr, from the number of moles we calculated from the given mass. \[{0.3929 KOH \over 0.3704HBr} = 1.0608\]
Finally, determine the limiting reagent from the two ratios. Since the actual ratio is higher, the limiting reagent must be HBr. If the ratio were smaller, the limiting reagent would have been KOH.

Answer 5

In the second step, shouldn't H_{2}O be 18?

Answer 6

\[H _{2}O\]

Answer 7

Ah yes. Sorry about that. It doesn't affect the final answer however.