$H_{2} (g) + C l_{2} (g) \to 2 H C l (g) +$ $H_2(g) + Cl_2(g) -> 2HCl(g)+$ . If 2 mol of hydrogen gas are mixed with 4 mol of chlorine gas, how many moles of hydrogen chloride will be produced?

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$H_{2} (g) + C l_{2} (g) \to 2 H C l (g) +$ $H_2(g) + Cl_2(g) -> 2HCl(g)+$ . If 2 mol of hydrogen gas are mixed with 4 mol of chlorine gas, how many moles of hydrogen chloride will be produced?

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Answer 1 · 2017-03-30T09:25:20

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 $H_{2}$ $H_2$ molecule reacts with 1 $C l_{2}$ $Cl_2$ molecule
Forming
2 $H C l$ $HCl$ 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 $6.022 \cdot 10^{23}$ $6.022*10^(23)$ . This number tells us how many particles there are. So in 1 mol there are $6.022 \cdot 10^{23}$ $6.022*10^(23)$ particles. This number is also referred to as the constant of Avogadro.

In the question the amount of moles for $H_{2}$ $H_2$ and for $C l_{2}$ $Cl_2$ are given. We have to calculate how much $H C l$ $HCl$ 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 $H C l$ $HCl$ will be produced.

We start of with the 2 moles of hydrogen gas. We know the following now:
1 mole of $H_{2}$ $H_2$ gives 2 moles of $H C l$ $HCl$
Therefore
2 moles of $H_{2}$ $H_2$ gives 4 moles of $H C l$ $HCl$

Now this same calculation can be done for the $C l_{2}$ $Cl_2$ :
1 mole of $C l_{2}$ $Cl_2$ reacts to 2 moles of $H C l$ $HCl$ , therefore
4 moles of $C l_{2}$ $Cl_2$ reacts to 8 moles of $H C l$ $HCl$ .

Now we see we have 2 different amount of moles for $H C l$ $HCl$ 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 $H_{2}$ $H_2$ , we can calculate with the ratio that:
2 moles of $H_{2}$ $H_2$ reacts with 2 moles of $C l_{2}$ $Cl_2$ .
We do have 2 moles of $C l_{2}$ $Cl_2$ available!

Let's see the other one
4 moles of $C l_{2}$ $Cl_2$ reacts with 4 moles of $H_{2}$ $H_2$
But we do not have 4 moles of $H_{2}$ $H_2$ !!

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

This situation can be stated as following: The $C l_{2}$ $Cl_2$ is abundant.

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$H_{2} (g) + C l_{2} (g) \to 2 H C l (g) +$ $H_2(g) + Cl_2(g) -> 2HCl(g)+$ . If 2 mol of hydrogen gas are mixed with 4 mol of chlorine gas, how many moles of hydrogen chloride will be produced?

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Explanation:

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H2(g)+Cl2(g)→2HCl(g)+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?

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$H_{2} (g) + C l_{2} (g) \to 2 H C l (g) +$ $H_2(g) + Cl_2(g) -> 2HCl(g)+$ . If 2 mol of hydrogen gas are mixed with 4 mol of chlorine gas, how many moles of hydrogen chloride will be produced?