The reaction of calcium with water is represented by the equation
`Ca+2H_(2)OtoCa(OH)_(2)+H_(2)`
What volume of `H_(2)`, at STP would be liberated when 8g of calcium completely reacts with water

Options

A. `4480 cm^(3)`
B. `2240 cm^(3)`
C. `1120 cm^(3)`
D. `0.4 cm^(3)`

Explanation

To solve the problem of how much hydrogen gas (H₂) is liberated when 8g of calcium (Ca) reacts with water, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction of calcium with water can be represented as:
\[ \text{Ca} + 2 \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2 \]

2. **Calculate the Molar Mass of Calcium**:
The molar mass of calcium (Ca) is approximately 40 g/mol.

3. **Calculate the Number of Moles of Calcium**:
Using the formula for moles:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
For 8g of calcium:
\[ n_{\text{Ca}} = \frac{8 \text{ g}}{40 \text{ g/mol}} = 0.2 \text{ moles} \]

4. **Determine the Moles of Hydrogen Gas Produced**:
From the balanced equation, we see that 1 mole of calcium produces 1 mole of hydrogen gas (H₂). Therefore:
\[ n_{\text{H}_2} = n_{\text{Ca}} = 0.2 \text{ moles} \]

5. **Calculate the Volume of Hydrogen Gas at STP**:
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters (or 22400 cm³). Thus, the volume of hydrogen gas produced is:
\[ V_{\text{H}_2} = n_{\text{H}_2} \times 22.4 \text{ L} \]
Converting moles to volume:
\[ V_{\text{H}_2} = 0.2 \text{ moles} \times 22.4 \text{ L/mole} = 4.48 \text{ L} \]
Converting liters to cm³:
\[ V_{\text{H}_2} = 4.48 \text{ L} \times 1000 \text{ cm}^3/\text{L} = 4480 \text{ cm}^3 \]

6. **Conclusion**:
The volume of hydrogen gas liberated when 8g of calcium reacts completely with water is:
\[ \boxed{4480 \text{ cm}^3} \]