The **reaction is given** as follows:

\[2Na(s) +2H_{2}O \Rightarrow 2NaOH(aq) +H_{2}(g)\]

This **article aims**Â to **determine the mass of sodium used in the given reaction. **The article uses the**Â concept of finding the number of moles**Â and then mass from the **number of moles. **Ideal gas law is used to determine the number of moles of gas. **Ideal gas law** relates **pressure**, **volume,** the **quantity of gas**, **temperature,** and the **number of moles**. The **ideal gas law** is given as:

\[PV = nRT \]

**Expert Answer**

**Given Data:**

The **reaction is given**Â as:

\[2Na(s) +2H_{2}O \Rightarrow 2NaOH(aq) +H_{2}(g)\]

$M_{r} = 22.99$

The **temperature in Kelvin**Â is given as:

$ T = 25 ^{\circ} C = (273+25)K = 298K $

$ V = 246mL = 0.246L $ is the volume of gas.

$ p_{t} = 1.00 \:atm$ is the **total pressure over the water.**

$ R = 0.0821 \: L\:atm \:mol^{-1}\: K^{-1} $ is the**Â ideal gas constant.**

$P_{H_{2}O} = 0.0313 \: atm $ at $ 25^{\circ} C $ is the **vapor pressure of water.**

The **partial pressure of the hydrogen**Â is given by:

\[p_{t} = p_{H_{2}}+ p_{H_{2}O}\]

\[1.00atm = p_{H_{2}} + 0.0313atm Â \]

\[p_{H_{2}} = 0.9687 atm\]

The**Â moles of hydrogen**Â are given by:

\[0.9687atm \times 0.246L = n \times 0.08206 L.atm.K^{-1}.mol^{-1} \times 298.15K \]

\[0.2383 = n\times 24.27 mol^{-1} \]

\[n =0.009740 mol\]

\[Moles\:of \: Na = 0.009740\: mol\:of \: H_{2} \times \dfrac{2 \:mol\: Na }{1\:mol\:H_{2}} = 0.01948molNa \]

\[Mass\: of \:Na = 0.01948molNa \times \dfrac {22.99\:g\:of \:Na }{1\:mol\:Na } = 0.449\:g\:of\:Na \]

The **mass**Â of $Na$ used was $0.449\:g$.

**Numerical Result**

The **mass of sodium used in the reaction**Â is $0.449\:g$.

**Example**

**A reaction of sodium metal with water is as follows:**

**\[2Na(s) +2H_{2}O \Rightarrow 2NaOH(aq) +H_{2}(g)\]**

**Hydrogen gas is gathered over water at $30.0 ^{\circ }C$. Volume of gas is $250\: mL$, measured at $1.00\: atm$. Determine the number of grams of sodium used in the reaction. (Water vapor pressure at $25$ degrees ATM.)**

**Solution**

**Given Data:**

The **reaction is given**Â below:

\[2Na(s) +2H_{2}O \Rightarrow 2NaOH(aq) +H_{2}(g)\]

$M_{r} = 22.99$

The **temperature in Kelvin**Â is given as:

$ T = 25 \circ C = (273+30)K = 303K $

$ V = 250mL = 0.25L $ is the volume of gas.

$ p_{t} = 1.00 \:atm$ is the **total pressure over the water.**

$ R = 0.0821 \: L\:atm \:mol^{-1}\: K^{-1} $ is the**Â ideal gas constant.**

$P_{H_{2}O} = 0.0313 \: atm $ at $ 25 ^{\circ} C $ is the **vapor pressure of water.**

The **partial pressure of the hydrogen**Â is given by:

\[p_{t} = p_{H_{2}}+ p_{H_{2}O}\]

\[1.00atm = p_{H_{2}} + 0.0313atm Â \]

\[p_{H_{2}} = 0.9687 atm\]

The**Â moles of hydrogen**Â are given by:

\[0.9687atm \times 0.25L = n \times 0.08206 L.atm.K^{-1}.mol^{-1} \times 303K \]

\[0.2422 = n\times 24.864 mol^{-1} \]

\[n =0.009741 mol\]

\[Moles\:of \: Na = 0.009741\: mol\:of \: H_{2} \times \dfrac{2 \:mol\: Na }{1\:mol\:H_{2}} = 0.019482molNa \]

\[Mass\: of \:Na = 0.019482molNa \times \dfrac {22.99\:g\:of \:Na }{1\:mol\:Na } = 0.449\:g\:of\:Na \]

The **mass**Â of $Na$ used was $0.4478\:g$.