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Solid sodium reacts with liquid water to form hydrogen gas according to the equation

2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)
What is the pressure (in atm) of hydrogen gas in the 20.0 L headspace of a reactor vessel when 3.36 kg sodium is reacted with excess water at 50.0 °C?

User Bonneville
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The pressure (in atm) of hydrogen gas in the 20.0 L headspace of a reactor vessel when 3.36 kg sodium is reacted with excess water at 50.0 °C can be calculated using the ideal gas law, which states that the pressure, volume, and temperature of a gas are related by the equation PV = nRT, where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas in Kelvin.

To calculate the pressure of the hydrogen gas, we first need to determine the number of moles of hydrogen gas produced by the reaction. We can do this by using the balanced chemical equation for the reaction, which is 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g).

From the balanced chemical equation, we can see that for every 2 moles of sodium that react, 1 mole of hydrogen gas is produced. Therefore, if 3.36 kg of sodium is reacted with excess water, the number of moles of hydrogen gas produced is 3.36 kg / (22.99 g/mol) = 146.4 moles.

Next, we need to convert the temperature of the gas from degrees Celsius to Kelvin. This can be done by adding 273.15 to the temperature in Celsius, which gives us a temperature of 50.0 + 273.15 = 323.15 K.

Now that we have the number of moles of hydrogen gas and the temperature of the gas, we can plug these values into the ideal gas law equation to calculate the pressure of the hydrogen gas:

P = (nRT) / V

P = (146.4 mol * 0.08206 L*atm / mol*K * 323.15 K) / 20.0 L

P = 4.31 atm

Therefore, the pressure (in atm) of hydrogen gas in the 20.0 L headspace of a reactor vessel when 3.36 kg sodium is reacted with excess water at 50.0 °C is 4.31 atm.

User LifeInstructor
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