Question
If 3.75 of zinc dust was added to excess tetraoxosulphate vi acid. Calculate the number of molecules of hydrogen produced (Zn=65)
Answers
GPT-4o mini
To calculate the number of molecules of hydrogen produced when zinc dust reacts with sulfuric acid (\( \text{H}_2\text{SO}_4 \)), we start with the balanced chemical equation for the reaction:
\[
\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2
\]
From the equation, we can see that one mole of zinc reacts with one mole of sulfuric acid to produce one mole of zinc sulfate and one mole of hydrogen gas.
1. **Calculate moles of zinc**:
Given the mass of zinc dust is 3.75 g and the molar mass of zinc (\( \text{Zn} \)) is 65 g/mol:
\[
\text{moles of Zn} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.75 \, \text{g}}{65 \, \text{g/mol}} \approx 0.0577 \, \text{mol}
\]
2. **Use the stoichiometry of the reaction**:
From the balanced reaction, we know that 1 mole of \( \text{Zn} \) produces 1 mole of \( \text{H}_2 \). Therefore, the moles of hydrogen produced will also be:
\[
\text{moles of } \text{H}_2 = 0.0577 \, \text{mol}
\]
3. **Convert moles of hydrogen to molecules**:
To find the number of molecules, we use Avogadro's number (\( 6.022 \times 10^{23} \) molecules/mol):
\[
\text{number of molecules of } \text{H}_2 = \text{moles} \times \text{Avogadro's number}
\]
\[
\text{number of molecules of } \text{H}_2 = 0.0577 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 3.47 \times 10^{22} \, \text{molecules}
\]
4. **Final answer**:
The number of molecules of hydrogen produced is approximately:
\[
3.47 \times 10^{22} \, \text{molecules of } \text{H}_2
\]
\[
\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2
\]
From the equation, we can see that one mole of zinc reacts with one mole of sulfuric acid to produce one mole of zinc sulfate and one mole of hydrogen gas.
1. **Calculate moles of zinc**:
Given the mass of zinc dust is 3.75 g and the molar mass of zinc (\( \text{Zn} \)) is 65 g/mol:
\[
\text{moles of Zn} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.75 \, \text{g}}{65 \, \text{g/mol}} \approx 0.0577 \, \text{mol}
\]
2. **Use the stoichiometry of the reaction**:
From the balanced reaction, we know that 1 mole of \( \text{Zn} \) produces 1 mole of \( \text{H}_2 \). Therefore, the moles of hydrogen produced will also be:
\[
\text{moles of } \text{H}_2 = 0.0577 \, \text{mol}
\]
3. **Convert moles of hydrogen to molecules**:
To find the number of molecules, we use Avogadro's number (\( 6.022 \times 10^{23} \) molecules/mol):
\[
\text{number of molecules of } \text{H}_2 = \text{moles} \times \text{Avogadro's number}
\]
\[
\text{number of molecules of } \text{H}_2 = 0.0577 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 3.47 \times 10^{22} \, \text{molecules}
\]
4. **Final answer**:
The number of molecules of hydrogen produced is approximately:
\[
3.47 \times 10^{22} \, \text{molecules of } \text{H}_2
\]