Question 36 of 36intermediate🔧 ApplyNumerical3 marks

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO₂(s) → 2H₂O (l) + MnCl₂(aq) + Cl₂(g) How many grams of HCl react with 5.0 g of manganese dioxide?

Correct Answer

8.4 g of HCl

Exercise: EXERCISES | Q: 1.36 | (Page 28)
For More Understanding

Explanation

To find the mass of HCl reacting with 5.0 g of MnO₂, we first convert the given mass of MnO₂ to moles using its molar mass. Then, using the stoichiometric ratio from the balanced chemical equation (4 moles of HCl react with 1 mole of MnO₂), we calculate the required moles of HCl. Finally, we convert the moles of HCl back to grams using the molar mass of HCl.

Solution Steps

  1. Step 1: Write the balanced equation and calculate molar masses

    4 HCl (aq)+MnO2(s)→2H2O (l)+MnCl2(aq)+Cl2(g)4 \text{ HCl (aq)} + \text{MnO}_2(\text{s}) \rightarrow 2\text{H}_2\text{O (l)} + \text{MnCl}_2(\text{aq}) + \text{Cl}_2(\text{g})

    Molar mass of MnO2\text{MnO}_2 = 55+2×16=87 g mol−155 + 2 \times 16 = 87 \text{ g mol}^{-1} Molar mass of HCl\text{HCl} = 1+35.5=36.5 g mol−11 + 35.5 = 36.5 \text{ g mol}^{-1}

  2. Step 2: Calculate moles of MnO2\text{MnO}_2 Moles of MnO2\text{MnO}_2 = Given massMolar mass=5.0 g87 g mol−1=0.0575 mol\frac{\text{Given mass}}{\text{Molar mass}} = \frac{5.0 \text{ g}}{87 \text{ g mol}^{-1}} = 0.0575 \text{ mol}

  3. Step 3: Determine moles of HCl required From the balanced equation, 1 mole of MnO2\text{MnO}_2 reacts with 4 moles of HCl. Moles of HCl = 4×0.0575 mol=0.23 mol4 \times 0.0575 \text{ mol} = 0.23 \text{ mol}

  4. Step 4: Calculate the mass of HCl Mass of HCl = Moles×Molar mass=0.23 mol×36.5 g mol−1=8.395 g\text{Moles} \times \text{Molar mass} = 0.23 \text{ mol} \times 36.5 \text{ g mol}^{-1} = 8.395 \text{ g} Therefore, approximately 8.4 g8.4 \text{ g} of HCl will react with 5.0 g5.0 \text{ g} of manganese dioxide.

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