Determine the theoretical yield of alum (MM=474.38), to two decimal places, if 0.36 g of aluminum (MM = 26.98 g/mol) are reacted. (Note: refer to the manual for the stoichiometric ratios)
- 冷眼旁觀Lv 61 month agoBest Answer
Each mole of alum (KAl(SO₄)₂•12H₂O) contains 1 mole of Al.
Stoichiometrically, each mole of Al can produce 1 mole of KAl(SO4)2•12H2O.
Moles of Al reacted = (0.36 g) / (26.98 g/mol) = 0.01334 mol
Maximum moles of KAl(SO₄)₂•12H₂O produced = 0.01334 mol
Theoretical yield of KAl(SO₄)₂•12H₂O = (0.01334 mol) × (474.38 g/mol) = 6.33 g
Theoretical yield of KAl(SO₄)₂•12H₂O
= (0.36 g Al) × (1 mol Al / 26.98 g Al) × (1 mol KAl(SO₄)₂•12H₂O / 1 mol Al) × (474.38 g KAl(SO₄)₂•12H₂O / 1 mol KAl(SO₄)₂•12H₂O)
= 6.33 g KAl(SO₄)₂•12H₂O
- Trevor HLv 71 month ago
Where is the missing data? Especially the mass of alum produced. Referring to your manual is of no value to us here.
- davidLv 71 month ago
(Note: refer to the manual for the stoichiometric ratios) <<< where is that damn manual? I just keep looking for my manual, but I can't find it.