To solve the problem of finding the volume ratio of hydrogen gas evolved when 5 g of zinc reacts with excess NaOH and dilute HCl, we can follow these steps:
### Step 1: Calculate Moles of Zinc
First, we need to calculate the number of moles of zinc (Zn) in 5 g.
\[
\text{Molar mass of Zn} = 65.38 \, \text{g/mol}
\]
\[
\text{Moles of Zn} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \, \text{g}}{65.38 \, \text{g/mol}} \approx 0.0764 \, \text{mol}
\]
### Step 2: Reaction with Excess NaOH
When zinc reacts with excess NaOH, it forms sodium zincate and hydrogen gas. The balanced chemical equation is:
\[
\text{Zn} + 2 \text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2
\]
From the equation, we see that 1 mole of zinc produces 1 mole of hydrogen gas.
### Step 3: Calculate Volume of Hydrogen from NaOH Reaction
At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of hydrogen gas produced from the reaction with NaOH is:
\[
\text{Volume of } H_2 = \text{moles of Zn} \times 22.4 \, \text{L/mol} = 0.0764 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 1.71 \, \text{L}
\]
### Step 4: Reaction with Dilute HCl
When zinc reacts with dilute HCl, the balanced chemical equation is:
\[
\text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
Again, 1 mole of zinc produces 1 mole of hydrogen gas.
### Step 5: Calculate Volume of Hydrogen from HCl Reaction
Using the same number of moles of zinc calculated earlier, the volume of hydrogen gas produced from the reaction with HCl is:
\[
\text{Volume of } H_2 = \text{moles of Zn} \times 22.4 \, \text{L/mol} = 0.0764 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 1.71 \, \text{L}
\]
### Step 6: Calculate Volume Ratio
Now, we can find the volume ratio of hydrogen gas evolved in both reactions:
\[
\text{Volume ratio} = \frac{\text{Volume of } H_2 \text{ from NaOH}}{\text{Volume of } H_2 \text{ from HCl}} = \frac{1.71 \, \text{L}}{1.71 \, \text{L}} = 1:1
\]
### Conclusion
The volume ratio of hydrogen gas evolved in reaction (I) with excess NaOH and reaction (II) with dilute HCl is **1:1**.
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