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**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 $0.4478\:g$.