Certainly! Let's work through this problem step by step.
### Step 1: Understanding the Reaction
The balanced chemical equation provided is:
[tex]\[ 4 HCl + MnO_2 \rightarrow 2 H_2O + MnCl_2 + Cl_2 \][/tex]
This tells us that 4 moles of HCl react with 1 mole of MnO_2 to produce the products.
### Step 2: Determine the Molar Masses
1. Molar Mass of MnO_2:
The atomic masses given are:
- Manganese (Mn) = 54.938 amu
- Oxygen (O) = 16 amu
Molar mass of MnO_2:
[tex]\[
\text{Molar mass of } MnO_2 = \text{Mass of Mn} + 2 \times \text{Mass of O}
\][/tex]
[tex]\[
\text{Molar mass of } MnO_2 = 54.938 + 2 \times 16 = 86.938 \, \text{g/mol}
\][/tex]
2. Molar Mass of HCl:
The atomic masses required are:
- Hydrogen (H) = 1 amu
- Chlorine (Cl) = 35.45 amu
Molar mass of HCl:
[tex]\[
\text{Molar mass of } HCl = \text{Mass of H} + \text{Mass of Cl}
\][/tex]
[tex]\[
\text{Molar mass of } HCl = 1 + 35.45 = 36.45 \, \text{g/mol}
\][/tex]
### Step 3: Calculate Moles of MnO_2
Given the mass of MnO_2 is 5 grams, we can calculate the number of moles:
[tex]\[
\text{Moles of } MnO_2 = \frac{\text{Mass of } MnO_2}{\text{Molar Mass of } MnO_2}
\][/tex]
[tex]\[
\text{Moles of } MnO_2 = \frac{5 \, \text{g}}{86.938 \, \text{g/mol}} \approx 0.057512 \, \text{moles}
\][/tex]
### Step 4: Relate Moles of HCl to Moles of MnO_2
From the balanced chemical equation, 1 mole of MnO_2 reacts with 4 moles of HCl. Therefore:
[tex]\[
\text{Moles of HCl} = 4 \times \text{Moles of } MnO_2
\][/tex]
[tex]\[
\text{Moles of HCl} = 4 \times 0.057512 \approx 0.230049 \, \text{moles}
\][/tex]
### Step 5: Calculate Mass of HCl Required
Finally, to find the mass of HCl that will react with the given MnO_2, we use the moles of HCl calculated:
[tex]\[
\text{Mass of HCl} = \text{Moles of HCl} \times \text{Molar Mass of HCl}
\][/tex]
[tex]\[
\text{Mass of HCl} = 0.230049 \, \text{moles} \times 36.45 \, \text{g/mol} \approx 8.385 \, \text{grams}
\][/tex]
### Final Answer
Approximately 8.385 grams of HCl are required to react with 5 grams of MnO_2.
