How much oxygen is needed to burn 6 g of aluminum containing 10% impurities.

1. Let’s write down the equation of aluminum combustion:

4Al + 3O2 = 2Al2O3;

2.Calculate the mass of aluminum in the sample:

m (Al) = w (Al) * m (sample);

w (Al) = 1 – w (impurities) = 1 – 0.1 = 0.9;

m (Al) = 0.9 * 6 = 5.4 g;

3. calculate the chemical amount of aluminum:

n (Al) = m (Al): M (Al) = 5.4: 27 = 0.2 mol;

4. find the amount of oxygen needed for the reaction:

n ((O2) = n (Al) * 3: 4 = 0.2 * 3: 4 = 0.15 mol;

5.determine the volume of oxygen:

V (O2) = n (O2) * Vm = 0.15 * 22.4 = 3.36 liters.

Answer: 3.36 liters.