Question
A piece of zinc was added to 100 cm³ of 0.2M hydrochloric acid. After effervescence head stopped, 30 cm³ of the resulting solution required 22 cm³ of a solution for complete neutralisation, calculate the mass of zinc added
Answers
GPT-4o mini
To solve this problem, we start by determining the reaction between zinc and hydrochloric acid. The balanced chemical equation for the reaction is:
\[
\text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
1. **Calculate the moles of hydrochloric acid initially present:**
The concentration of hydrochloric acid is \(0.2 \, \text{M}\) and the volume is \(100 \, \text{cm}^3\):
\[
\text{Volume in L} = \frac{100 \, \text{cm}^3}{1000} = 0.1 \, \text{L}
\]
\[
\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol}
\]
2. **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 \(C \times 0.022\):
\[
\text{Moles of HCl remaining in 30 cm³} = C \times 0.022
\]
3. **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,
\[
0.02 - (C \cdot 0.022) \text{ moles}
\]
4. **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:
\[
\text{Moles of Zn} = \frac{\text{Moles of HCl reacted}}{2}
\]
5. **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.
\[
\text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
1. **Calculate the moles of hydrochloric acid initially present:**
The concentration of hydrochloric acid is \(0.2 \, \text{M}\) and the volume is \(100 \, \text{cm}^3\):
\[
\text{Volume in L} = \frac{100 \, \text{cm}^3}{1000} = 0.1 \, \text{L}
\]
\[
\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol}
\]
2. **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 \(C \times 0.022\):
\[
\text{Moles of HCl remaining in 30 cm³} = C \times 0.022
\]
3. **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,
\[
0.02 - (C \cdot 0.022) \text{ moles}
\]
4. **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:
\[
\text{Moles of Zn} = \frac{\text{Moles of HCl reacted}}{2}
\]
5. **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.