Use PV = nRT and solve for n = number of moles oxygen. Then use this link as an example for solving stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
4Al + 3O2 = 2Al2O3
It takes 2.00 L pure oxygen gas at STP (0C ,1atm ) to react completely with a certain aluminum. What is the mass of aluminum reacted?
http://www.jiskha.com/science/chemistry/stoichiometry.html
Let's follow these steps:
Step 1: Convert the given volume of oxygen to moles.
Given:
- Volume of oxygen gas (V) = 2.00 L
- Temperature (T) = 0°C (which is 273.15 K at STP)
- Pressure (P) = 1 atm
At STP, 1 mole of any gas occupies 22.4 L. Using the ideal gas law, we can convert the volume of oxygen gas to moles:
n = PV/RT
where:
- n is the number of moles
- P is the pressure in atm
- V is the volume in liters
- R is the ideal gas constant (0.0821 Lâ‹…atm/molâ‹…K)
- T is the temperature in Kelvin
Plugging in the values:
n = (1 atm) * (2.00 L) / (0.0821 Lâ‹…atm/molâ‹…K * 273.15 K)
Step 2: Determine the stoichiometric ratio between aluminum (Al) and oxygen (O2) in the balanced equation.
From the balanced equation:
4Al + 3O2 --> 2Al2O3
The stoichiometric ratio between aluminum and oxygen is 4:3. This means that for every 4 moles of aluminum, 3 moles of oxygen are required.
Step 3: Calculate the moles of aluminum reacted.
Since the stoichiometric ratio is 4:3, the moles of aluminum reacted will be (4/3) times the moles of oxygen reacted. Multiply the number of moles of oxygen obtained in Step 1 by the stoichiometric ratio:
moles of aluminum = (4/3) * moles of oxygen
Step 4: Convert moles of aluminum reacted to mass.
The molar mass of aluminum (Al) is approximately 26.98 g/mol. Multiply the moles of aluminum by the molar mass to obtain the mass of aluminum reacted:
mass of aluminum = moles of aluminum * molar mass of aluminum
By following these steps, you can calculate the mass of aluminum reacted using the given volume of oxygen gas at STP.