Consider the following reaction

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

The mass will be 12.73 g ?!!!!

aswear

To determine the mass of aluminum reacted, we need to use stoichiometry to relate the volume of oxygen to the moles of aluminum, and then convert moles to mass using the molar mass of aluminum.

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 L⋅atm/mol⋅K)
- T is the temperature in Kelvin

Plugging in the values:
n = (1 atm) * (2.00 L) / (0.0821 L⋅atm/mol⋅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.