Question

How many grams of HCl will completely react with 17.4g of pure MnO\(_2\) (s) to liberate Cl\(_2\) (g)? (Atomic mass Mn = 55.0; H = 1; Cl = 35.5)

• A. 14.6 g
• B. 7.3 g
• C. 21.9 g
• D. 29.2 g
• E. 34.8 g

Hint:

To solve stoichiometric problems, always start by balancing the equation, then calculate the moles of reactants and products involved, and use molar masses to find the desired quantity.

Answer 1

The Correct Option is D

The reaction for the liberation of chlorine gas from MnO\(_2\) is given by: \[ {MnO}_2 (s) + 4 {HCl} (aq) \rightarrow {MnCl}_2 (aq) + {Cl}_2 (g) + 2 {H}_2{O} (l) \] From the balanced equation, we see that 1 mole of MnO\(_2\) reacts with 4 moles of HCl.
Now, let’s calculate the moles of MnO\(_2\) in 17.4 g: \[ {Moles of MnO}_2 = \frac{17.4}{55.0 + 2(16)} = \frac{17.4}{87.0} = 0.2 \, {mol} \] Since 1 mole of MnO\(_2\) reacts with 4 moles of HCl, the moles of HCl required are: \[ {Moles of HCl} = 0.2 \times 4 = 0.8 \, {mol} \] Now, we calculate the mass of HCl needed: \[ {Mass of HCl} = 0.8 \, {mol} \times (1 + 35.5) = 0.8 \times 36.5 = 29.2 \, {g} \] Thus, the required mass of HCl is 29.2 g.