CHEMISTRY HIGH SCHOOL

A piece of sodium metal reacts completely with water as follows: 2Na(s) + 2H2O(l) ⟶ 2NaOH(aq) + H2(g) The hydrogen gas generated is collected over water at 25.0°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°C = 0.0313 atm.)

Answers

Answer 1
Answer:

Answer:

\large \boxed{\text{0.449 g}}

Explanation:

1. Gather all the information in one place

M_r:   22.99

           2Na + 2H₂O ⟶ 2NaOH + H₂

\, p_{\text{tot}} = \quad \text{1.00 atm}\np_{\text{H2O}} = \text{0.0313 atm}

T = 25.0 °C

V = 246 mL

2. Moles of H₂

To find the moles of hydrogen, we can use the Ideal Gas Law:

pV = nRT

(a) Calculate the partial pressure of the hydrogen

p_{\text{tot}} = p_{\text{H2}} + p_{\text{H2O}}\n\text{1.00 atm} =p_{\text{H2}} +\text{0.0313 atm}\np_{\text{H2}} = \text{0.9687 atm}

(b) Convert the volume to litres

V = 246 mL = 0.246 L

(c) Convert the temperature to kelvins

T = (25.0 + 273.15) K = 298.15 K

(d) Calculate the moles of hydrogen

\begin{array}{rcl}\text{0.9687 atm}* \text{0.246 L} & = & n * 0.082 06 \text{ L}\cdot\text{atm}\cdot\text{K}^(-1)\text{mol}^(-1) * \text{298.15 K}\n0.2383 & = & 24.47n \text{ mol}^(-1)\n\nn & = & \frac{0.2383}{24.47\text{ mol}^(-1)}\n\n& = & 0.009740 \text{ mol}\n\end{array}

3. Moles of Na

The molar ratio is 2 mol Na: 1 mol H₂

\text{Moles of Na} =\text{0.009 740 mol H}_(2) * \frac{\text{2 mol Na}}{\text{1 mol H}_(2)} = \text{0.019 48 mol Na}

4. Mass of Na

\text{Mass of Na} = \text{0.019 48 mol Na} * \frac{\text{22.99 g Na}}{\text{1 mol Na}} = \text{0.449 g Na}\n\n\text{The mass of Na used was $\large \boxed{\textbf{0.449 g}}$}
Answer 2
Answer:

The number of grams of sodium used in the reaction will be approximately 0.1387 grams.

To calculate the number of grams of sodium used in the reaction, we need to use the ideal gas law and consider the effect of the vapor pressure of water on the pressure of the collected hydrogen gas.

Given data:

Volume of hydrogen gas (V) = 246 mL = 0.246 L

Pressure of hydrogen gas (P) = 1.00 atm

Vapor pressure of water (P_water vapor) = 0.0313 atm (subtracted from the total pressure)

Temperature (T) = 25.0°C = 298.15 K

The ideal gas law is given by the equation: PV = nRT, where n is the number of moles of the gas.

First, calculate the total pressure by subtracting the vapor pressure of water from the given pressure of the gas:

Total pressure (P_total) = P - P_water vapor

= 1.00 atm - 0.0313 atm = 0.9687 atm

Now, rearrange the ideal gas law equation to solve for n (moles of gas):

n = PV / RT

Plug in the values:

n = (0.9687 atm × 0.246 L) / (0.0821 L·atm/mol·K × 298.15 K)

n ≈ 0.01206 mol

According to the balanced chemical equation, 2 moles of sodium (Na) produce 1 mole of hydrogen gas (H₂). Therefore, the number of moles of sodium used in the reaction is half of the calculated moles of hydrogen gas:

Moles of sodium = 0.01206 mol / 2 = 0.00603 mol

Finally, calculate the mass of sodium (molar mass of sodium = 22.99 g/mol)

Mass of sodium = Moles of sodium × Molar mass of sodium

Mass of sodium = 0.00603 mol × 22.99 g/mol

≈ 0.1387 g

Therefore, the number of grams of sodium used in the reaction is approximately 0.1387 grams.

To know more about sodium here

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