Limiting STOICH - Please show work, chem is hard.

Question

What mass of Aluminum oxide is formed when 13.0g of Aluminum is burned in 21.0g of Oxygen gas?

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Answer 1

First, you must write the correctly balanced equation for the reaction taking place:

4Al(s) + 3O₂(g) ==> 2Al₂O₃(s) ... balanced equation

Next, find the limiting reactant. One easy way is to divide the moles of each reactant by the corresponding coefficient in the balanced equation.

Al: 13.0 g Al x 1 mol Al / 26.98 g = 0.482 mols Al (÷4 -> 0.12)

O2: 21.0 g O₂ x 1 mol O₂ / 32 g = 0.656 mols O₂ (÷3 -> 0.219)

Al is LIMITING since 0.12 is less than 0.219

Finally, compute the mass of Al₂O₃ using the limiting reactant, stoichiometry and dimensional analysis:

0.482 mols Al x 2 mol Al₂O₃ / 4 mol Al x 102 g Al₂O₃ / mol = 24.6 g Al₂O₃ produced