How many grams of barium hydride must be treated with water to obtain 4.36L of hydrogen at `20^(@)C` and 0.975 atm pressure (Ba=137)?

Answer

To solve the problem of how many grams of barium hydride (BaH₂) must be treated with water to obtain 4.36 L of hydrogen gas (H₂) at 20°C and 0.975 atm pressure, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between barium hydride and water can be represented as:
\[ \text{BaH}_2 + 2 \text{H}_2\text{O} \rightarrow \text{Ba(OH)}_2 + \text{H}_2 \]

### Step 2: Use the Ideal Gas Law to find moles of hydrogen gas
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure in atm (0.975 atm)
- \( V \) = volume in liters (4.36 L)
- \( n \) = 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 equation to solve for \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ n = \frac{(0.975 \, \text{atm}) \times (4.36 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (293 \, \text{K})} \]

Calculating:
\[ n = \frac{4.247 \, \text{atm·L}}{24.1063 \, \text{L·atm/(K·mol)}} \]
\[ n \approx 0.1767 \, \text{moles of } H_2 \]

### Step 3: Relate moles of hydrogen to moles of barium hydride
From the balanced equation, we see that 1 mole of BaH₂ produces 1 mole of H₂. Therefore, the moles of BaH₂ needed can be calculated as:
\[ \text{moles of BaH}_2 = \frac{\text{moles of } H_2}{2} \]
Since 1 mole of BaH₂ produces 1 mole of H₂, we have:
\[ \text{moles of BaH}_2 = \frac{0.1767}{2} = 0.08835 \, \text{moles of BaH}_2 \]

### Step 4: Calculate the mass of barium hydride required
The molar mass of barium hydride (BaH₂) can be calculated as follows:
- Molar mass of Ba = 137 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of BaH₂ = 137 + (2 × 1) = 139 g/mol

Now, we can calculate the mass of BaH₂ needed:
\[ \text{mass of BaH}_2 = \text{moles of BaH}_2 \times \text{molar mass of BaH}_2 \]
\[ \text{mass of BaH}_2 = 0.08835 \, \text{moles} \times 137 \, \text{g/mol} \]
\[ \text{mass of BaH}_2 \approx 12.1039 \, \text{grams} \]

### Final Answer
The mass of barium hydride required is approximately **12.1039 grams**.

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