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

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the balanced chemical equation
The reaction between aluminum (Al) and caustic soda (NaOH) in the presence of water (H2O) can be represented as:
\[ 2 \text{Al} + 2 \text{NaOH} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaAlO}_2 + 3 \text{H}_2 \]
From this equation, we can see that 2 moles of aluminum produce 3 moles of hydrogen gas (H2).

### Step 2: Calculate the moles of aluminum
To find the moles of aluminum in 0.15 g, we use the formula:
\[ \text{Moles of Al} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of aluminum (Al) is approximately 27 g/mol. Thus:
\[ \text{Moles of Al} = \frac{0.15 \text{ g}}{27 \text{ g/mol}} \approx 5.56 \times 10^{-3} \text{ moles} \]

### Step 3: Calculate the moles of hydrogen produced
From the balanced equation, we know that 2 moles of Al produce 3 moles of H2. Therefore, the moles of H2 produced can be calculated as follows:
\[ \text{Moles of H}_2 = \left(\frac{3}{2}\right) \times \text{Moles of Al} \]
Substituting the value of moles of Al:
\[ \text{Moles of H}_2 = \left(\frac{3}{2}\right) \times 5.56 \times 10^{-3} \approx 8.33 \times 10^{-3} \text{ moles} \]

### Step 4: Calculate the volume of hydrogen gas produced
Using the ideal gas law, we can find the volume of hydrogen gas produced at the given conditions (20°C and 1 bar). The ideal gas equation is:
\[ PV = nRT \]
Where:
- \( P \) = pressure (1 bar = 100 kPa)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0831 L·bar/(K·mol))
- \( T \) = temperature in Kelvin (20°C = 293 K)

Rearranging the equation to solve for volume \( V \):
\[ V = \frac{nRT}{P} \]
Substituting the known values:
\[ V = \frac{(8.33 \times 10^{-3} \text{ moles}) \times (0.0831 \text{ L·bar/(K·mol)}) \times (293 \text{ K})}{1 \text{ bar}} \]
Calculating this gives:
\[ V \approx 0.203 \text{ L} \]

### Final Answer
The volume of hydrogen gas released when 0.15 g of aluminum reacts is approximately **0.203 liters**.

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