What volume of hydrogen gas at STP is produced by the complete reaction of 15.0 g of sodium according to the following equation:
2Na + 2H2O -----> 2NaOH + H2?
The balanced equation shows that 2 moles of sodium produce 1 mole of hydrogen gas.
First, we need to calculate the number of moles of sodium in 15.0 g:
moles of Na = mass / molar mass = 15.0 g / 22.99 g/mol = 0.652 moles
Since 2 moles of Na produce 1 mole of H2, we can use stoichiometry to calculate the moles of H2 produced:
moles of H2 = moles of Na / 2 = 0.652 moles / 2 = 0.326 moles
Finally, we can use the ideal gas law to find the volume of H2 at STP (standard temperature and pressure, which is 0°C and 1 atm):
PV = nRT
where P = 1 atm, V is the volume we want to find, n = 0.326 moles, R is the gas constant (0.08206 L atm/mol K), and T = 273 K.
Solving for V:
V = (nRT) / P = (0.326 mol * 0.08206 L atm/mol K * 273 K) / 1 atm = 6.90 L
Therefore, 15.0 g of sodium at STP produces 6.90 L of hydrogen gas.
To find the volume of hydrogen gas produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (which is 1 atm at STP)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L*atm/mol*K)
T = temperature (which is 273 K at STP)
First, we need to find the number of moles of sodium (Na) in 15.0 g. To do this, we divide the mass by the molar mass of sodium.
Molar mass of Na = 22.99 g/mol
Number of moles of Na = 15.0 g / 22.99 g/mol = 0.652 mol
According to the balanced equation, 1 mole of sodium (Na) produces 1 mole of hydrogen gas (H2).
So, the number of moles of H2 produced = 0.652 mol
Now, we can use the ideal gas law to find the volume of H2 by rearranging the equation:
V = (nRT) / P
V = (0.652 mol * 0.0821 L*atm/mol*K * 273 K) / 1 atm
V = 14.9 L
Therefore, the volume of hydrogen gas produced at STP by the complete reaction of 15.0 g of sodium is 14.9 L.