Final answer:
To calculate the volume of hydrogen gas evolved from the reaction of 23g of sodium with an excess of water, you need to balance the chemical equation and use the ideal gas law. From the balanced equation, you can determine the number of moles of hydrogen gas produced and then use the ideal gas law to calculate the volume. The volume of hydrogen gas at STP is 11.21 L.
Step-by-step explanation:
To calculate the volume of hydrogen gas evolved from the reaction of 23g of sodium with an excess of water, we need to balance the chemical equation first. The balanced equation for the reaction is:
2Na + 2H2O -> 2NaOH + H2
From the balanced equation, we can see that 2 moles of sodium (Na) react with 2 moles of water (H2O) to produce 1 mole of hydrogen gas (H2). Therefore, the number of moles of hydrogen gas produced can be calculated by converting the mass of sodium (23g) to moles, using its molar mass (23g/mol).
Using the formula:
Number of moles = Mass / Molar mass
We can calculate:
Number of moles of sodium = 23g / 23g/mol = 1 mole
Number of moles of hydrogen gas = 1 mole / 2 = 0.5 moles
Now, to calculate the volume of hydrogen gas at STP, we can use the ideal gas law:
PV = nRT
Where P = pressure (STP = 1 atm), V = volume, n = number of moles (0.5 moles), R = ideal gas constant (0.0821 L.atm/mol.K), and T = temperature (STP = 273 K).
Substituting the values into the equation, we can calculate the volume of hydrogen gas:
V = (nRT) / P = (0.5 moles * 0.0821 L.atm/mol.K * 273 K) / 1 atm = 11.21 L