[tex]$
4 \text{HCl (aq)} + \text{MnO}_2 \text{(s)} \longrightarrow \text{MnCl}_2 \text{(aq)} + 2 \text{H}_2\text{O (g)} + \text{Cl}_2\text{(g)}
$[/tex]

Chlorine gas can be made in the lab by the reaction of hydrochloric acid and manganese(IV) oxide, as shown in the reaction above. How many moles of chlorine can form from 8.98 moles of hydrochloric acid?



Answer :

Certainly! Let's solve this step-by-step:

1. Understand the balanced chemical equation:
[tex]\[ 4 \text{HCl (aq)} + \text{MnO}_2 \text{ (s)} \rightarrow \text{MnCl}_2 \text{ (aq)} + 2 \text{H}_2\text{O (g)} + \text{Cl}_2 \text{ (g)} \][/tex]
This equation tells us that 4 moles of hydrochloric acid (HCl) react with manganese(IV) oxide (MnO₂) to produce manganese(II) chloride (MnCl₂), water (H₂O), and chlorine gas (Cl₂).

2. Identify the stoichiometric relationship:
From the balanced equation, we see that 4 moles of HCl produce 1 mole of Cl₂. This is the stoichiometric ratio we will use for our calculations.

3. Determine the amount of chlorine gas produced:
We are given 8.98 moles of HCl and want to find out how many moles of Cl₂ are formed.

4. Apply the stoichiometric ratio:
Using the ratio from the balanced equation, we see:
[tex]\[ \frac{4 \text{ moles HCl}}{1 \text{ mole Cl}_2} \][/tex]
We can set up a proportion to find the moles of Cl₂ produced:
[tex]\[ \text{Moles of Cl}_2 = \frac{\text{Moles of HCl}}{4} \][/tex]

5. Substitute the given amount of HCl into the proportion:
[tex]\[ \text{Moles of Cl}_2 = \frac{8.98 \text{ moles HCl}}{4} \][/tex]

Simplifying this:
[tex]\[ \text{Moles of Cl}_2 = 2.245 \text{ moles} \][/tex]

So, 2.245 moles of chlorine gas (Cl₂) can be formed from 8.98 moles of hydrochloric acid (HCl).