Answer:
35.1 g
Explanation:
Given data:
Mass of Al(OH)₃ = 100 g
Mass of water formed = ?
Solution:
Chemical equation:
2Al(OH)₃ → Al₂O₃ + 3H₂O
Number of moles of Al(OH)₃:
Number of moles = mass/molar mass
Number of moles = 100 g/ 78 g/mol
Number of moles = 1.3 mol
now we will compare the moles of Al(OH)₃ with water.
Al(OH)₃ : H₂O
2 : 3
1.3 : 3/2×1.3 = 1.95
Mass of water produced:
Mass = number of moles ×molar mass
Mass = 1.95 mol × 18 g/mol
Mass = 35.1 g
Answer: 
concentration of
= 0.50 M
concentration of
= 0.25 M
Explanation:
The dissociation equation of
is:

According to stoichiometry:
1 mole of
gives 2 moles of 
Thus 0.25 moles of
gives =
moles of 
Similarly,
1 mole of
gives = 1 mole of 
Thus 0.25 moles of
gives =
moles of 
Thus the concentration of
and
are 0.50 M and 0.25 M respectively.
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