Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4HCl(aq) + MnO2(l) + MnCl2(aq)+Cl2(g)

Correct Option: The correct answer is 8.4g.

Understanding the Question:
The question asks us to calculate the mass of hydrochloric acid (HCl) required to react completely with 5.0 g of manganese dioxide (MnO₂) according to the balanced chemical equation:

$$4\mathrm{HCl}(\mathrm{aq})+{\mathrm{MnO}}_2$$

Step 1: Calculate the molar masses of the reactants
First, we find the molar mass of manganese dioxide (MnO₂) and hydrochloric acid (HCl):
• Molar mass of Mn = 54.94 g/mol
• Molar mass of O = 16.00 g/mol
• Molar mass of H = 1.008 g/mol
• Molar mass of Cl = 35.45 g/mol

$$\text{Molar mass of}{\mathrm{MnO}}_2=54.94+(2\times 16.00)=86.94\text{g/mol}$$

$$\text{Molar mass of}\mathrm{HCl}=1.008+35.45=36.46\text{g/mol}$$

Step 2: Determine stoichiometric relationship
According to the balanced chemical equation, 1 mole of MnO₂ reacts completely with 4 moles of HCl.
Expressing this in terms of mass:
• Mass of 1 mole of MnO₂ = 86.94 g
• Mass of 4 moles of HCl = $4\times 36.46\text{g}=145.84\text{g}$

Step 3: Calculate the mass of HCl required for 5.0 g of MnO₂
Using the unitary method, we calculate the amount of HCl needed to react with 5.0 g of MnO₂:

$$\text{Mass of}\mathrm{HCl}=(\frac{145.84\text{g}}{}$$
86.94 g ) × 5.0 g

$$\text{Mass of}\mathrm{HCl}\approx 1.6775\times 5.0\text{g}\approx 8.388\text{g}$$

Rounding to two significant figures, we get 8.4g of HCl. Therefore, 8.4 g of hydrochloric acid is required to react with 5.0 g of manganese dioxide.