The drain cleaner Drainex contains small bits of aluminium which react with caustic soda to produce hydrogen What volume of hydrogen at `20^(@)C` and one bar will be released when `0.15 g` of aluminium reacts ? .

Explanation

To find the volume of hydrogen gas released when 0.15 g of aluminum reacts with caustic soda, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction of aluminum with caustic soda (sodium hydroxide) can be represented as:
\[ 2 \text{Al (s)} + 2 \text{NaOH (aq)} + 2 \text{H}_2\text{O (l)} \rightarrow 2 \text{NaAlO}_2 \text{(aq)} + 3 \text{H}_2 \text{(g)} \]

### Step 2: Calculate the moles of aluminum
To find the number of moles of aluminum, we use the formula:
\[ \text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \]
Given:
- Mass of aluminum = 0.15 g
- Molar mass of aluminum = 27 g/mol

Calculating the moles:
\[ \text{Moles of Al} = \frac{0.15 \, \text{g}}{27 \, \text{g/mol}} = 5.56 \times 10^{-3} \, \text{mol} \]

### Step 3: Use stoichiometry to find moles of hydrogen produced
From the balanced equation, we see that 2 moles of aluminum produce 3 moles of hydrogen gas. Therefore, the moles of hydrogen produced can be calculated as:
\[ \text{Moles of H}_2 = \frac{3}{2} \times \text{Moles of Al} \]
Substituting the moles of aluminum:
\[ \text{Moles of H}_2 = \frac{3}{2} \times 5.56 \times 10^{-3} = 8.34 \times 10^{-3} \, \text{mol} \]

### Step 4: Calculate the volume of hydrogen gas using the ideal gas law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (1 bar = 1 atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (20°C = 293 K)

Rearranging the ideal gas law to find volume:
\[ V = \frac{nRT}{P} \]
Substituting the known values:
\[ V = \frac{(8.34 \times 10^{-3} \, \text{mol}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (293 \, \text{K})}{1 \, \text{atm}} \]

Calculating the volume:
\[ V \approx \frac{(8.34 \times 10^{-3}) \times (0.0821) \times (293)}{1} \approx 0.203 \, \text{L} \]

### Step 5: Convert volume to milliliters
To convert liters to milliliters:
\[ V = 0.203 \, \text{L} \times 1000 \, \text{mL/L} = 203 \, \text{mL} \]

### Final Answer
The volume of hydrogen gas released when 0.15 g of aluminum reacts with caustic soda is approximately **203 mL**.

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