Question

Chlorine is prepared in the laboratory by treating manganese dioxide \((MnO_2)\) with aqueous hydrochloric acid according to the reaction
\(4 HCl (aq) + MnO_2(s) → 2H_2O (l) + MnCl_2(aq) + Cl_2 (g)\).
How many grams of HCl react with 5.0 g of manganese dioxide?

Answer 1

Given:

Mass of MnO₂ = 5.0 g
Molar mass of MnO₂ = 86.94 g mol⁻¹
Molar mass of HCl = 36.46 g mol⁻¹

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Step 1: Write the balanced chemical equation

4HCl(aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g)

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Step 2: Calculate moles of MnO₂

Number of moles of MnO₂ = Mass / Molar mass

= 5.0 / 86.94

= 0.0575 mol

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Step 3: Calculate moles of HCl required

From the equation,
1 mol MnO₂ reacts with 4 mol HCl

Moles of HCl required = 4 × 0.0575

= 0.23 mol

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Step 4: Calculate mass of HCl

Mass of HCl = Moles × Molar mass

= 0.23 × 36.46

= 8.38 g

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Final Answer:

Mass of HCl required to react with 5.0 g of MnO₂
= 8.38 g