Answer:
1. The limiting reactant is H₂O.
2. The theoretical yield of Al₂O₃ = 978.8 g.
Step-by-step explanation:
1. Which is the limiting reactant?
- The balanced equation for the mentioned reaction is:
2Al + 3H₂O → Al₂O₃ + 3H₂,
It is clear that 2.0 moles of Al react with 3.0 moles of H₂O to produce 1.0 moles of Al₂O₃ and 3.0 mole of H₂.
- From stichiometry; Al reacts with H₂O with (2: 3) molar ratio.
∴ 4.8 mol of Al (the remaining 1.7 mol is in excess) reacts completely with 7.2 mol of H₂O with (2: 3) molar ratio.
the limiting reactant is H₂O and the excess reactant is Al.
2. Calculate the theoretical yield.
- To calculate the theoretical yield, we should get the no. of moles of produced Al₂O₃:
Using cross multiplication:
2.0 moles of Al produce → 1.0 moles of Al₂O₃.
4.8 moles of Al produce → ??? moles of Al₂O₃.
∴ The no. of moles of Al₂O₃ produced = (1.0 mol)(4.8 mol)/(2.0 mol) = 9.6 mol.
∴ The theoretical yield of Al₂O₃ = (no. of moles)(molar mass) = (9.6 mol)(101.96 g/mol) = 978.8 g.