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?
a
14.6 L
b
10.2 L
c
7.30 L
d
18.3 L
To solve this problem, we need to use stoichiometry and the ideal gas law. Here are the steps:
1. Write the balanced chemical equation: 2Na + 2H2O → 2NaOH + H2
2. Find the moles of Na from its mass:
moles of Na = 15.0 g / 23.0 g/mol = 0.652 moles of Na
3. Use the mole ratio between Na and H2 to find the moles of H2:
moles of H2 = 0.652 moles of Na × (1 mole of H2 / 2 moles of Na) = 0.326 moles of H2
4. Use the ideal gas law to find the volume of H2 at STP (Standard Temperature and Pressure):
PV = nRT, where P = 1 atm, V = ?, n = 0.326 moles, R = 0.0821 L·atm/(mol·K), T = 273 K.
V = nRT/P = (0.326 moles) × (0.0821 L·atm/(mol·K)) × (273 K)/(1 atm) ≈ 7.3 L
Therefore, the answer is (c) 7.30 L.
To determine the volume of hydrogen gas produced at STP (Standard Temperature and Pressure) by the complete reaction of 15.0 g of sodium, we need to use the stoichiometry of the balanced equation.
The coefficient in front of the hydrogen gas (H2) indicates that 2 moles of hydrogen gas are produced for every 2 moles of sodium reacted. Therefore, the molar ratio is 1:1.
Now, let's calculate the moles of sodium:
Molar mass of sodium (Na) = 22.99 g/mol
Moles of sodium = mass of sodium / molar mass of sodium
Moles of sodium = 15.0 g / 22.99 g/mol
Moles of sodium = 0.652 mol
Since the molar ratio between hydrogen gas and sodium is 1:1, the moles of hydrogen gas produced will be the same as the moles of sodium.
Now, let's use Avogadro's law to calculate the volume of gas at STP.
1 mole of any ideal gas at STP occupies 22.4 L.
Volume of hydrogen gas (H2) = Moles of hydrogen gas * 22.4 L/mol
Volume of hydrogen gas = 0.652 mol * 22.4 L/mol
Volume of hydrogen gas = 14.6 L
Therefore, the volume of hydrogen gas at STP produced by the complete reaction of 15.0 g of sodium is 14.6 L.
So, the correct answer is a) 14.6 L.