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.
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