Answer:
40.3 grams
Step-by-step explanation:
The balanced chemical equation for the reaction between aluminum (Al) and oxygen (O2) to form aluminum oxide (Al2O3) is:
4Al + 3O2 → 2Al2O3
To determine the amount of aluminum oxide produced, we need to first identify the limiting reactant, which is the reactant that gets completely used up in the reaction.
Using the given masses of Al and O2, we can calculate the number of moles of each:
moles of Al = 10.0 g / 26.98 g/mol = 0.371 mol
moles of O2 = 19.0 g / 32.00 g/mol = 0.594 mol
To determine the limiting reactant, we need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation. The stoichiometric coefficient of Al is 4, while the coefficient of O2 is 3. Therefore, O2 is the limiting reactant because it produces fewer moles of Al2O3 than the amount of Al available.
The moles of Al2O3 produced can be calculated using the mole ratio between O2 and Al2O3 from the balanced equation:
moles of Al2O3 = (0.594 mol O2) x (2 mol Al2O3 / 3 mol O2) = 0.396 mol Al2O3
Finally, we can convert the moles of Al2O3 produced to grams using its molar mass:
mass of Al2O3 = 0.396 mol x 101.96 g/mol = 40.3 g
Therefore, 40.3 grams of aluminum oxide are produced in the reaction.