B) calculate the ratio of moles of h2o to moles of anhydrous kal(so4)2. note: report the ratio to the closest whole number. (sho
1 answer:
Answer: -
Mass of Hydrated KAl(SO₄)₂ = 2.0 g
Molar mass of anhydrous KAl(SO₄)₂ = 258.20 g/ mol
Mass of of anhydrous KAl(SO₄)₂ = mass of the 2nd heating = A
The mass of water released = mass of the Aluminum Cup + 2.0 grams of KAl(SO₄)₂ - mass of the 2nd heating
= H g
Moles of water released =
Moles of anhydrous KAl(SO₄)₂ =
Required ratio =
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Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of = 183.511 g/mole
- First we have to calculate the moles of Cu.
The moles of Cu = 4.7209 moles
From the given chemical formula, we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of = 4.4209 moles
- Now we have to calculate the mass of
.
Mass of = Moles of
× Molar mass of
= 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.