2 Na + 2 H2O -> 2 NaOH + H2
First, we need to calculate the moles of Na involved in the reaction. The molar mass of Na is 23 g/mol.
Therefore, moles of Na = 20 g / 23 g/mol = 0.87 mol
From the balanced equation, we can see that 2 moles of Na react to produce 1 mole of H2 gas.
Therefore, moles of H2 produced = 0.87 mol / 2 = 0.435 mol
Next, we need to calculate the volume of H2 gas produced using the ideal gas law equation:
PV = nRT
Where:
P = 104 kPa = 104000 Pa
V = volume of gas in liters
n = moles of gas = 0.435 mol
R = ideal gas constant = 8.31 L*kPa/mol*K
T = temperature in Kelvin = 300 K
Rearranging the equation to solve for V:
V = (nRT) / P
V = (0.435 mol * 8.31 L*kPa/mol*K * 300 K) / 104000 Pa
V = 1.025 L
Therefore, 1.025 liters of hydrogen gas can be produced at 300 K and 104 kPa when 20 grams of sodium metal is reacted with water.
use ideal gas law to figure out. how many liters of hydrogen gas can be produced of hydrogen gas can be produced at 300 k and 104 kpa if 20 grams of sodium metal is reaced with water according to the following equation
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