Yellowish-green gas chlorine `(Cl_(2))` can be prepared in the laboratory by heating hydrochloric acid `(HCl, aq)` with pyrolusite (manganese dioxide, `MnO_(2)`):
`4HCl(aq.) + MnO_(2)(s) rarr Cl_(2)(g) + 2H_(2)O(l) + MnCl_(2)(aq.)`.
How many grams of `HCl` reacts with `5.00g` of manganese dioxide?

Explanation

To determine how many grams of hydrochloric acid (HCl) react with 5.00 grams of manganese dioxide (MnO₂), we will follow these steps:

### Step 1: Calculate the molar mass of MnO₂
The molar mass of MnO₂ can be calculated by adding the atomic masses of manganese (Mn) and oxygen (O):
- Manganese (Mn) = 55 g/mol
- Oxygen (O) = 16 g/mol (and there are 2 oxygen atoms in MnO₂)

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

### Step 2: Calculate the number of moles of MnO₂
Using the mass of MnO₂ given (5.00 g), we can calculate the number of moles of MnO₂:

\[ \text{Number of moles of MnO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \text{ g}}{87 \text{ g/mol}} \approx 0.0575 \text{ moles} \]

### Step 3: Use the stoichiometry of the reaction
From the balanced chemical equation:

\[ 4 \text{HCl} + \text{MnO}_2 \rightarrow \text{Cl}_2 + 2 \text{H}_2\text{O} + \text{MnCl}_2 \]

We see that 1 mole of MnO₂ reacts with 4 moles of HCl. Therefore, the number of moles of HCl required for 0.0575 moles of MnO₂ is:

\[ \text{Moles of HCl} = 4 \times \text{moles of MnO}_2 = 4 \times 0.0575 \approx 0.230 \text{ moles} \]

### Step 4: Calculate the mass of HCl
Now, we need to calculate the mass of HCl using its molar mass. The molar mass of HCl is approximately 36.5 g/mol.

\[ \text{Mass of HCl} = \text{moles of HCl} \times \text{molar mass of HCl} = 0.230 \text{ moles} \times 36.5 \text{ g/mol} \approx 8.395 \text{ g} \]

### Step 5: Round the answer
Rounding to two decimal places, we find that approximately 8.40 grams of HCl are required to react with 5.00 grams of MnO₂.

### Final Answer
**8.40 grams of HCl react with 5.00 grams of MnO₂.**

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