A small piece of zinc is dissolved in \(50.00 \mathrm{mL}\) of \(1.035 \mathrm{M}\) HCl. At the conclusion of the reaction, the concentration of the \(50.00 \mathrm{mL}\) sample is redetermined and found to be \(0.812 \mathrm{M} \mathrm{HCl} .\) What must have been the mass of the piece of zinc that dissolved? $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

Short Answer

Expert verified
The mass of the piece of zinc that dissolved is approximately \(0.364 \, \text{grams}\).

Step by step solution

01

Calculate Initial Moles of Hydrochloric Acid

The initial moles of hydrochloric acid are obtained by multiplying the initial concentration by the volume (Expressed in litres). Which equals to \(1.035 \, \text{M} \times 0.05 \, \text{L} = 0.05175 \, \text{mol}\).
02

Calculate Final Moles of Hydrochloric Acid

Similar to the first step, calculate the final moles of hydrochloric acid. Multiply the final concentration by the volume: \(0.812 \, \text{M} \times 0.05 \, \text{L} = 0.0406 \, \text{mol}\).
03

Calculate Moles of Zinc

The reaction between zinc and hydrochloric acid is a 1:2 reaction, meaning 1 mole of zinc reacts with 2 moles of hydrochloric acid. So, the moles of zinc that reacted would be half of the moles of hydrochloric acid that reacted. Therefore, subtract the final moles of HCl from the initial moles to get the moles of HCl that reacted and divide by 2 to get the moles of zinc that reacted: \((0.05175 - 0.0406) \, \text{mol} / 2 = 0.005575 \, \text{mol}\).
04

Find Mass of Zinc

Finally, convert moles of zinc to grams using the molar mass of zinc (65.38 g/mol): \(0.005575 \, \text{mol} \times 65.38 \, \text{g/mol} = 0.364 \, \text{g}\).

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Most popular questions from this chapter

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