How many gm of HCl is needed for complete reaction with 43.5 gm `MnO_(2)` ? (Mn = 55)

Answer

To determine how many grams of HCl are needed for the complete reaction with 43.5 grams of MnO₂, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction between MnO₂ and HCl can be represented as:
\[ \text{MnO}_2 + 4 \text{HCl} \rightarrow \text{MnCl}_2 + 2 \text{H}_2\text{O} + \text{Cl}_2 \]

### Step 2: Determine the molar mass of MnO₂.
The molar mass of MnO₂ can be calculated as follows:
- Molar mass of Mn = 55 g/mol
- Molar mass of O = 16 g/mol
- Therefore, molar mass of MnO₂ = 55 + (2 × 16) = 55 + 32 = 87 g/mol

### Step 3: Calculate the number of moles of MnO₂ in 43.5 grams.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
We can find the number of moles of MnO₂:
\[ \text{Number of moles of MnO}_2 = \frac{43.5 \text{ g}}{87 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 4: Use the stoichiometry of the reaction to find moles of HCl required.
From the balanced equation, we see that 1 mole of MnO₂ reacts with 4 moles of HCl. Therefore, for 0.5 moles of MnO₂:
\[ \text{Moles of HCl} = 0.5 \text{ moles MnO}_2 \times 4 \frac{\text{moles HCl}}{\text{mole MnO}_2} = 2 \text{ moles HCl} \]

### Step 5: Calculate the mass of HCl required.
The molar mass of HCl is 36.5 g/mol. Thus, the mass of HCl required can be calculated as:
\[ \text{Mass of HCl} = \text{moles of HCl} \times \text{molar mass of HCl} \]
\[ \text{Mass of HCl} = 2 \text{ moles} \times 36.5 \text{ g/mol} = 73 \text{ g} \]

### Final Answer:
Therefore, the mass of HCl needed for the complete reaction with 43.5 grams of MnO₂ is **73 grams**.

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