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 a complete reaction with 43.5 grams of MnO₂, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrochloric acid (HCl) and manganese dioxide (MnO₂) can be represented as:
\[ \text{HCl} + \text{MnO}_2 \rightarrow \text{MnCl}_2 + \text{H}_2\text{O} + \text{Cl}_2 \]
Balancing the equation gives:
\[ 4 \text{HCl} + \text{MnO}_2 \rightarrow \text{MnCl}_2 + 2 \text{H}_2\text{O} + \text{Cl}_2 \]

### Step 2: Determine the molar masses
- Molar mass of Mn = 55 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of Cl = 35.5 g/mol

Calculating the molar mass of MnO₂:
\[ \text{Molar mass of MnO}_2 = 55 + (2 \times 16) = 55 + 32 = 87 \text{ g/mol} \]

Calculating the molar mass of HCl:
\[ \text{Molar mass of HCl} = 1 + 35.5 = 36.5 \text{ g/mol} \]

### Step 3: Use stoichiometry to find the required grams of HCl
From the balanced equation, we see that 4 moles of HCl react with 1 mole of MnO₂.

Using the molar masses:
- 87 g of MnO₂ reacts with 146 g of HCl (since \(4 \times 36.5 = 146\)).

### Step 4: Set up a proportion to find grams of HCl needed for 43.5 g of MnO₂
Using the ratio:
\[ \frac{146 \text{ g HCl}}{87 \text{ g MnO}_2} = \frac{x \text{ g HCl}}{43.5 \text{ g MnO}_2} \]

Cross-multiplying gives:
\[ x = \frac{146 \times 43.5}{87} \]

### Step 5: Calculate x
Calculating the right side:
\[ x = \frac{146 \times 43.5}{87} = \frac{6351}{87} \approx 73 \text{ g HCl} \]

### Conclusion
Thus, **73 grams of HCl** are needed for the complete reaction with 43.5 grams of MnO₂.

---