**reaction is given**as follows: \[2Na(s) +2H_{2}O \Rightarrow 2NaOH(aq) +H_{2}(g)\] This

**article aims**to

**determine the mass of sodium used in the given reaction.**The article uses the

**concept of finding the number of moles**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**as: \[2Na(s) +2H_{2}O \Rightarrow 2NaOH(aq) +H_{2}(g)\] $M_{r} = 22.99$ The

**temperature in Kelvin**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

**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**is given by: \[p_{t} = p_{H_{2}}+ p_{H_{2}O}\] \[1.00atm = p_{H_{2}} + 0.0313atm \] \[p_{H_{2}} = 0.9687 atm\] The

**moles of hydrogen**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**of $Na$ used was $0.449\:g$.

**Numerical Result**

The **mass of sodium used in the reaction**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**below: \[2Na(s) +2H_{2}O \Rightarrow 2NaOH(aq) +H_{2}(g)\] $M_{r} = 22.99$ The

**temperature in Kelvin**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

**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**is given by: \[p_{t} = p_{H_{2}}+ p_{H_{2}O}\] \[1.00atm = p_{H_{2}} + 0.0313atm \] \[p_{H_{2}} = 0.9687 atm\] The

**moles of hydrogen**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**of $Na$ used was $0.4478\:g$.