banner

A piece of sodium metal reacts completely with water is given below. The hydrogen gas generated is collected over water at 25.0 degree C. The volume of the gas is 246 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25 degree C=0.0313 atm.)

The 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 $0.4478\:g$.

5/5 - (5 votes)