Consider the following chemical reaction.
2H2O(l) -->2H2(g)+O2(g)
What mass of H2O is required to form 1.6 L of O2 at a temperature of 330 K and a pressure of 0.954 atm?
To determine the mass of H2O required to form 1.6 L of O2 at a given temperature and pressure, we need to use the ideal gas law equation.
The ideal gas law equation is given by:
PV = nRT
Where:
- P is the pressure of the gas (in atmospheres, atm)
- V is the volume of the gas (in liters, L)
- n is the number of moles of the gas
- R is the ideal gas constant (0.0821 L·atm/mol·K)
- T is the temperature in Kelvin (K)
First, let's convert the given temperature from Celsius to Kelvin:
T = 330 K
Next, we'll rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Now we can substitute the given values into the equation to calculate the number of moles:
n = (0.954 atm) * (1.6 L) / (0.0821 L·atm/mol·K * 330 K)
n = 0.0741 moles of O2
According to the balanced chemical equation, the stoichiometric ratio between O2 and H2O is 1:2. This means that for every 1 mole of O2, we need 2 moles of H2O.
Using this ratio, we can calculate the number of moles of H2O required:
n(H2O) = 2 * n(O2)
n(H2O) = 2 * 0.0741 moles
n(H2O) = 0.1482 moles
Finally, we'll calculate the mass of H2O using its molar mass, which is approximately 18.02 g/mol:
mass(H2O) = n(H2O) * molar mass(H2O)
mass(H2O) = 0.1482 moles * 18.02 g/mol
mass(H2O) ≈ 2.67 g
Therefore, approximately 2.67 grams of H2O is required to form 1.6 L of O2 at a temperature of 330 K and a pressure of 0.954 atm.
example.
http://www.jiskha.com/science/chemistry/stoichiometry.html