How many liters of hydrogen gas can be produced at 300.0K and 1.55atm if 20.0g of sodium metal is reacted with water?
See your post above.
To find the number of liters of hydrogen gas produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, let's determine the number of moles of hydrogen gas produced from the reaction of 20.0g of sodium metal with water.
The balanced chemical equation for the reaction between sodium and water is:
2 Na + 2 H2O → 2 NaOH + H2
From the equation, we can see that 2 moles of sodium react to produce 1 mole of hydrogen gas.
To find the number of moles of sodium, we can use the molar mass of sodium:
Mass of sodium (Na) = 20.0 g/mol
Molar mass of sodium (Na) = 22.99 g/mol
Now, we calculate the number of moles of sodium:
n = (mass of sodium) / (molar mass of sodium)
n = 20.0 g / 22.99 g/mol
n ≈ 0.870 mol
Since 2 moles of sodium react to produce 1 mole of hydrogen gas, the number of moles of hydrogen gas produced is half of the number of moles of sodium:
Number of moles of hydrogen gas = 0.870 mol / 2
Number of moles of hydrogen gas = 0.435 mol
Now, let's substitute the known values into the ideal gas law equation to find the volume of hydrogen gas:
PV = nRT
V = (nRT) / P
V = (0.435 mol) * (0.0821 L.atm/mol.K) * (300.0 K) / 1.55 atm
V ≈ 5.88 L
Therefore, approximately 5.88 liters of hydrogen gas can be produced at 300.0K and 1.55atm when 20.0g of sodium metal is reacted with water.