To solve this problem, we start by determining the reaction between zinc and hydrochloric acid. The balanced chemical equation for the reaction is:
\[ \text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]
-
Calculate the moles of hydrochloric acid initially present:
The concentration of hydrochloric acid is \(0.2 , \text{M}\) and the volume is \(100 , \text{cm}^3\):
\[ \text{Volume in L} = \frac{100 , \text{cm}^3}{1000} = 0.1 , \text{L} \]
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.2 , \text{mol/L} \times 0.1 , \text{L} = 0.02 , \text{mol} \]
-
Determine the remaining HCl after the reaction with zinc:
After the reaction, we are told that 30 cm³ of the resulting solution required 22 cm³ of another solution for complete neutralization.
Let’s assume this solution is a NaOH solution (for ease of calculation). We will first express the 30 cm³ in liters:
\[ \text{Volume in L} = \frac{30 , \text{cm}^3}{1000} = 0.03 , \text{L} \]
Now, the volume of the neutralizing solution required is 22 cm³, also converting to liters:
\[ \text{Volume in L} = \frac{22 , \text{cm}^3}{1000} = 0.022 , \text{L} \]
We assume the NaOH solution has a concentration of \(C\) M:
\[ \text{Moles of NaOH} = C \times 0.022 , \text{L} \]
For neutralization reactions between strong acids and bases, they react in a 1:1 ratio. Therefore, the moles of HCl present in the 30 cm³ that reacted with the NaOH will also equal \(C \times 0.022\):
\[ \text{Moles of HCl remaining in 30 cm³} = C \times 0.022 \]
-
Calculate the moles of HCl that reacted with the zinc:
The original moles of HCl was 0.02 mol, and the moles of HCl in the 30 cm³ solution was:
\[ \text{Moles of HCl in 30 cm³} = C \times 0.022 \]
To find how much HCl reacted with zinc, we can find the moles of HCl that were in the 100 cm³ but not present in the 30 cm³:
\[ \text{Moles of HCl reacted} = \text{Original moles} - \text{Remaining moles} \]
Since we know that remaining moles are:
\[ \text{Total moles} = \text{Moles in remaining 30 cm³} \]
Thus,
\[ 0.02 - (C \cdot 0.022) \text{ moles} \]
-
Use the moles of HCl that reacted to find the moles of zinc:
Since the reaction is 1 mole of Zn for 2 moles of HCl, we can relate:
\[ \text{Moles of Zn} = \frac{\text{Moles of HCl reacted}}{2} \]
-
Calculate the mass of zinc:
We first need to determine \(C\). If we have a certain concentration of NaOH (for example, 1M), we can substitute \(C = 1\):
\[ \text{Remaining moles HCl} = C \cdot 0.022 = 1 \cdot 0.022 = 0.022 , \text{mol} \]
Therefore:
\[ \text{Moles of HCl that reacted} = 0.02 - 0.022 = -0.002 , \text{mol} \quad (\text{this indicates a wrong assumption for C}) \]
This should lead us to readjust the assumed concentration C of the neutralizing solution (use backtrack to see viable concentrations).
For precise calculations, examine via an assumed realistic concentration from your knowledge of solutions.
Finally, if all calculations align, with C determined:
\[ \text{Mass of Zn} = \text{Moles of Zn} \times 65.38, \text{g/mol} \]
This will yield the mass of zinc added. If refining stepwise clarity doesn't yield an answer, adjust to compute remaining moles via trials in neutralization using solvable context for C accordingly.
1 answer