Consider the following reaction.

4 Al(s) + 3 O2(g) → 2 Al2O3(s)
It takes 1.00 L pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

Just follow the steps in this worked example of a stoichiometry problem.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To find the mass of aluminum reacted, we need to use the stoichiometry of the balanced chemical equation and the molar mass of aluminum.

1. Determine the number of moles of oxygen gas using the ideal gas law equation:

PV = nRT

Where:
P = pressure (STP = 1 atm)
V = volume of gas (1.00 L)
n = number of moles (to be determined)
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (STP = 273 K)

Using the equation:

(1 atm)(1.00 L) = n(0.0821 L.atm/mol.K)(273 K)

Solving for n, we find:

n = 0.0409 moles

2. Use the stoichiometry of the balanced equation to find the moles of aluminum reacted. According to the balanced equation, the ratio of aluminum to oxygen is 4:3. Therefore, the moles of aluminum reacted would be:

(4/3) * 0.0409 moles = 0.0545 moles

3. Calculate the mass of aluminum using the molar mass of aluminum:

The molar mass of aluminum (Al) is 26.98 g/mol.

Mass of aluminum = moles of aluminum * molar mass of aluminum
= 0.0545 moles * 26.98 g/mol
≈ 1.47 grams

Therefore, the mass of aluminum reacted is approximately 1.47 grams.

To determine the mass of aluminum reacted, we need to use stoichiometry, which involves using the balanced chemical equation and the molar ratios between reactants and products.

Step 1: Convert the given volume of oxygen to moles.
Since the reaction is at standard temperature and pressure (STP), we can use the molar volume of a gas at STP, which is 22.4 L/mol. In this case, we have 1.00 L of oxygen gas. So, we can calculate the number of moles of oxygen using the following conversion:
1.00 L O2 × (1 mol / 22.4 L) = 0.0446 mol O2

Step 2: Use the stoichiometric ratio to find the moles of aluminum.
According to the balanced equation, the ratio of moles of oxygen to aluminum is 3:4. Therefore, for every 3 moles of oxygen, we have 4 moles of aluminum. Using this ratio, we can calculate the moles of aluminum as follows:
0.0446 mol O2 × (4 mol Al / 3 mol O2) = 0.0595 mol Al

Step 3: Calculate the mass of aluminum using its molar mass.
The molar mass of aluminum (Al) is 26.98 g/mol. Multiplying the moles of aluminum by its molar mass gives us the mass of aluminum reacted:
0.0595 mol Al × 26.98 g/mol = 1.60 g

Therefore, the mass of aluminum reacted is 1.60 grams.