# C is given and the solution Calculate the volume of oxygen required for the combustion

**C is given and the solution Calculate the volume of oxygen required for the combustion of 2.7 g of aluminum. (This reaction occurs when the sparklers are lit.)**

Let’s execute the solution:

1. Let us write the equation according to the problem statement:

4Al + 3O2 = 2Al2O3 – combustion of aluminum in oxygen, aluminum oxide is obtained;

2. Calculation of the molar masses of substances:

M (Al) = 26.9 g / mol;

M (O2) = 32 g / mol.

3. Determine the amount of moles of aluminum, oxygen:

Y (Al) = m / M = 2.7 / 26.9 = 0.1 mol;

0.1 mol (Al) – X mol (O2);

-4 mol – 3 mol from here, X mol (O2) = 0.1 * 3/4 = 0.075 mol.

4. Find the volume of O2, applying Avogadro’s law:

V (O2) = 0.075 * 22.4 = 1.68 liters.

Answer: 1.68 liters are required to burn aluminum. oxygen.