2Na + 2H2O = 2NaOH + H2
Calculat the moles and volume of hydrogen produced when 0.2 moles of sodium reacts with excess water.
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the moles and volume of hydrogen produced, we need to use the balanced chemical equation:
2Na + 2H2O --> 2NaOH + H2
From the equation, we can see that 2 moles of sodium (2Na) react with 2 moles of water (2H2O) to produce 1 mole of hydrogen gas (H2).
Given that 0.2 moles of sodium (Na) reacts with excess water (H2O), we can calculate the moles and volume of hydrogen produced.
Moles of hydrogen produced = (0.2 moles of sodium) x (1 mole of H2 / 2 moles of Na)
= 0.1 moles of H2
Since one mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters, we can calculate the volume of hydrogen gas produced.
Volume of hydrogen gas produced = (0.1 moles of H2) x (22.4 L)
= 2.24 L
Therefore, when 0.2 moles of sodium reacts with excess water, 0.1 moles of hydrogen gas is produced, which is equivalent to 2.24 liters of hydrogen gas.
To calculate the moles and volume of hydrogen produced, we will use the balanced chemical equation you provided:
2Na + 2H2O = 2NaOH + H2
First, we need to determine the stoichiometry of the reaction. From the equation, we can see that 2 moles of sodium (2Na) react with 2 moles of water (2H2O) to produce 1 mole of hydrogen gas (H2).
Since we have 0.2 moles of sodium (Na), we will use stoichiometry to determine the moles and volume of hydrogen produced.
Moles of H2 = (0.2 moles of Na) × (1 mole of H2 / 2 moles of Na)
Moles of H2 = 0.1 moles
Therefore, 0.2 moles of sodium reacting with excess water will produce 0.1 moles of hydrogen.
Next, to calculate the volume 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 = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
However, the temperature and pressure are not given in the question. Assuming standard temperature and pressure (STP), the temperature is 273 K, and the pressure is 1 atm.
Using the ideal gas law, we can rearrange the equation to solve for V:
V = (nRT) / P
Substituting the values, we have:
V = (0.1 moles) × (0.0821 L·atm/(mol·K)) × (273 K) / (1 atm)
V ≈ 2.24 L
Therefore, when 0.2 moles of sodium reacts with excess water under standard conditions, approximately 0.1 moles of hydrogen gas are produced, occupying a volume of approximately 2.24 liters.