To determine the number of moles of Al₂O₃ produced from 17.0 L of O₂ at STP, we use stoichiometry. 0.808 moles of Al₂O₃ will be produced.
To determine the number of moles of Al₂O₃ produced from 17.0 L of O₂ at STP, we need to use stoichiometry. First, we need to balance the equation for the reaction:
4 Al + 3 O₂ → 2 Al₂O₃
From the balanced equation, we can see that 4 moles of Al react with 3 moles of O₂ to produce 2 moles of Al₂O₃. Therefore, we can set up a proportion:
(2 moles of Al₂O₃ / 3 moles of O₂) = (x moles of Al₂O₃ / 17.0 L of O₂)
Simplifying the proportion, we can solve for x, which represents the number of moles of Al₂O₃:
x = (2/3) x (17.0 / 22.414)
x = 0.808 moles of Al2O3