please answer and explain
更新1:
9*10^22 sorry >>> EDIT
Chemistry question?
2017-12-18 12:20 am
5.4 g of aluminum is reacted with 9 1022 molecules of oxygen. How many milligrams of aluminum oxide formed in the reaction?
please answer and explain
please answer and explain
回答 (1)
2017-12-18 4:36 am
Initial no. of moles of Al = (5.4 g) / (27 g/mol) = 0.2 mol
Initial no. of moles of O₂ = (9 × 10²² molecules) / (6 × 10²³ molecules/mol) = 0.15 mol
Balanced equation for the reaction :
4Al + 3O₂ → 2Al₂O₃
Mole ratio Al : O₂ : Al₂O₃ = 4 : 3 : 2
If 0.2 mol Al completely reacts, O₂ required = (0.2 mol) × (3/4) = 0.15 mol = Initial no. of moles of O₂
Hence, both Al and O₂ completely react.
No. of moles of Al₂O₃ formed = (0.2 mol) × (2/4) = 0.1 mol
[OR: No. of moles of Al₂O₃ formed = (0.15 mol) × (2/3) = 0.1 mol]
Mass of Al₂O₃ formed = (0.1 mol) × (102 g/mol) = 10.2 g = 10200 mg
Initial no. of moles of O₂ = (9 × 10²² molecules) / (6 × 10²³ molecules/mol) = 0.15 mol
Balanced equation for the reaction :
4Al + 3O₂ → 2Al₂O₃
Mole ratio Al : O₂ : Al₂O₃ = 4 : 3 : 2
If 0.2 mol Al completely reacts, O₂ required = (0.2 mol) × (3/4) = 0.15 mol = Initial no. of moles of O₂
Hence, both Al and O₂ completely react.
No. of moles of Al₂O₃ formed = (0.2 mol) × (2/4) = 0.1 mol
[OR: No. of moles of Al₂O₃ formed = (0.15 mol) × (2/3) = 0.1 mol]
Mass of Al₂O₃ formed = (0.1 mol) × (102 g/mol) = 10.2 g = 10200 mg
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