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).
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).