How many liters of hydrogen gas can be produced at 300.0K and 1.55atm if 20.0g of sodium metal is reacted with water?
2Na + 2H2O ==> 2NaOH + H2
mols Na = grams/molar mass = ?
mols H2 = 1/2 that (look at the coefficients in the balanced equation).
Now use PV = nRT to convert mols to L at the conditions listed.
To solve this question, we need to use stoichiometry and the ideal gas law equation.
First, let's write the balanced chemical equation for the reaction between sodium and water:
2 Na + 2 H2O -> 2 NaOH + H2
This equation tells us that each mole of sodium (Na) reacts with 2 moles of water (H2O) to produce 1 mole of hydrogen gas (H2).
Step 1: Calculate moles of sodium (Na)
Given the mass of sodium is 20.0g, we need to convert it to moles using the molar mass of sodium (22.99 g/mol):
moles of Na = mass of Na / molar mass of Na
moles of Na = 20.0 g / 22.99 g/mol
moles of Na = 0.870 mol
Step 2: Calculate moles of hydrogen gas (H2)
Since the stoichiometry tells us that 1 mole of sodium produces 1 mole of hydrogen gas, the moles of hydrogen gas will be the same as the moles of sodium:
moles of H2 = moles of Na
moles of H2 = 0.870 mol
Step 3: Convert moles of hydrogen gas to volume in liters
Now, we can use the ideal gas law equation (PV = nRT), where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
We are given the following values:
P = 1.55 atm
T = 300.0 K
n = 0.870 mol
R = 0.0821 L·atm/mol·K
To find V, we rearrange the ideal gas law equation:
V = (nRT) / P
V = (0.870 mol * 0.0821 L·atm/mol·K * 300.0 K) / 1.55 atm
V ≈ 42.1 L
Therefore, approximately 42.1 liters of hydrogen gas can be produced at 300.0K and 1.55atm when 20.0g of sodium metal reacts with water.