When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is ________ . (Nearest integer)
Given :
Molar mass of Al is 27.0 g mol⁻¹
Molar mass of O is 16.0 g mol⁻¹

Solution

Chemistry ma’am · 4 steps

Convert both masses to moles first, then let the balanced equation tell you which one runs out.

1. Balanced equation. 4Al + 3O₂ → 2Al₂O₃. Every 4 mol of Al uses 3 mol of O₂ and gives 2 mol of Al₂O₃.
2. Moles available. Al: 81.0/27.0 = 3 mol. O₂ has molar mass 32.0 g mol⁻¹, so 128.0/32.0 = 4 mol.
3. Find the limiting reagent. 3 mol of Al needs only 3 × 3/4 = 2.25 mol of O₂. There are 4 mol, so oxygen is in excess and Al is the limiting reagent.

   n(Al₂O₃) = 3 × 2/4 = 1.5 mol

4. So. Molar mass of Al₂O₃ is 2 × 27.0 + 3 × 16.0 = 102 g mol⁻¹, so the mass formed is 1.5 × 102 = 153 g.

The common slip: 272. Treating O₂ as the limiting reagent gives 4 × 2/3 × 102 = 272 g, but only 2.25 mol of the 4 mol of O₂ can actually react.

The idea to keep. Never compare grams to find the limiting reagent. Divide each reactant's moles by its coefficient; the smallest ratio runs out first and fixes the product.

Answer from the NTA final answer key. Solution by PrepWiser.