# H_2(g) + Cl_2(g) -> 2HCl(g)+. If 2 mol of hydrogen gas are mixed with 4 mol of chlorine gas, how many moles of hydrogen chloride will be produced?

Mar 30, 2017

#### Answer:

4 moles of HCl are produced

#### Explanation:

The ratio at which the compounds react with each other are displayed in a chemical reaction as the numbers in front of the compounds. For example in this reaction:

1 ${H}_{2}$ molecule reacts with 1 $C {l}_{2}$ molecule
Forming
2 $H C l$ molecules.

In chemistry moles are used as a handy number of a certain amount of stuff. You can compare this to a dozen eggs, which is always 12. In the same way 1 mol is the number $6.022 \cdot {10}^{23}$. This number tells us how many particles there are. So in 1 mol there are $6.022 \cdot {10}^{23}$ particles. This number is also referred to as the constant of Avogadro.

In the question the amount of moles for ${H}_{2}$ and for $C {l}_{2}$ are given. We have to calculate how much $H C l$ is produced with both these values. Therefore 2 calculations needs to be done.

Because moles just tells us something about the amount of particles we can use the ratio of the reaction to calculate how many $H C l$ will be produced.

We start of with the 2 moles of hydrogen gas. We know the following now:
1 mole of ${H}_{2}$ gives 2 moles of $H C l$
Therefore
2 moles of ${H}_{2}$ gives 4 moles of $H C l$

Now this same calculation can be done for the $C {l}_{2}$:
1 mole of $C {l}_{2}$ reacts to 2 moles of $H C l$, therefore
4 moles of $C {l}_{2}$ reacts to 8 moles of $H C l$.

Now we see we have 2 different amount of moles for $H C l$ and the question now is which one should we use?
We must now look at which one of above reactions can occur? So for the 2 moles of ${H}_{2}$, we can calculate with the ratio that:
2 moles of ${H}_{2}$ reacts with 2 moles of $C {l}_{2}$.
We do have 2 moles of $C {l}_{2}$ available!

Let's see the other one
4 moles of $C {l}_{2}$ reacts with 4 moles of ${H}_{2}$
But we do not have 4 moles of ${H}_{2}$!!

Therefore we can only make 4 moles of $H C l$ instead of 8 moles.

This situation can be stated as following: The $C {l}_{2}$ is abundant.