# Question #916e4

Oct 28, 2014

12.5 L of ${H}_{2} O$ is produced.

4 mol HCl react with 1 mol ${O}_{2}$
So 4L HCl react with 1L ${O}_{2}$
So 50L ${O}_{2}$ require 4 x 50 = 200L HCl
We only have 25L of HCl so the ${O}_{2}$ is in excess.
Since 4L HCl gives 2L ${H}_{2} O$:
25 L HCl must give 25/2 = 12.5L ${H}_{2} O$

Oct 29, 2014

The limiting reagent is HCl, which means the reaction can only produce $\text{12.5 L H"_2"O}$.

#### Explanation:

In order to determine the volume of water produced when 25.0 L HCl reacts with 50.0 L ${\text{O}}_{2}$, the limiting reactant (sometimes called the limiting reagent ) must be calculated.

Balanced equation:

$\text{4HCl(g)" + "O"_2("g")}$$\rightarrow$$\text{2Cl"_2"(g)" + "2H"_2"O(g)}$

Since all substances are gases, the amount of each substance can be kept as liters. There is no need to convert liters to moles.

Determine the limiting reactant.
Multiply the known volume of each reactant times the mole ratio from the equation that leaves water and cancels $\text{HCl}$ or ${\text{O}}_{2}$.

$25.0 \cancel{\text{L HCl"xx (2 "L H"_2"O")/(4 cancel"L HCl")="12.5 L H"_2"O}}$

$50.0 \cancel{\text{L O"_2xx(2 "L H"_2"O")/(1 cancel"L O"_2)= "100 L H"_2"O}}$

$\text{HCl}$ is the limiting reactant because it produces less water than $\text{O"_2}$, therefore, the volume of water produced by the above reaction will produce $\text{12.5 xL H"_2"O(g)}$.