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Answer:
0.200L of 0.300 Al(OH)₃ are necessaries
Explanation:
A diprotic acid, H₂X, reacts with aluminium (III) hydroxide, Al(OH)₃ as follows:
3 H₂X + 2 Al(OH)₃ → 6H₂O + Al₂X₃
To solve this question we must find the moles of the H₂X, using the balanced reaction we can find the moles of Al(OH)₃ and its volumen knowing its molar concentration is 0.300M:
Moles H₂X:
0.300L * (0.300mol / L) = 0.0900moles H₂X
Moles Al(OH)₃:
0.0900moles H₂X * (2mol Al(OH)₃ / 3mol H₂X) = 0.0600 moles Al(OH)₃
Volume 0.300M Al(OH)₃:
0.0600 moles Al(OH)₃ * (1L / 0.300moles) =
The volume of Al(OH)₃ needed for the reaction is 0.2 L
We'll begin by writing the balanced equation for the reaction.
3H₂X + 2Al(OH)₃ → 6H₂O + Al₂X₃
The mole ratio of the acid, H₂X (nA) = 3
The mole ratio of base, Al(OH)₃ (nB) = 2
From the question given above, the following data were
Volume of acid, H₂X (Va) = 0.3 L
Molarity of acid, H₂X (Ma) = 0.3 M
Molarity of base, Al(OH)₃ (Mb) = 0.3 M
MaVa / MbVb = nA / nB
(0.3 × 0.3) / (0.3 × Vb) = 3/2
0.09 / (0.3 × Vb) = 3/2
Cross multiply
0.09 × 2 = 0.3 × 3 × Vb
0.18 = 0.9 × Vb
Divide both side by 0.9
Vb = 0.18 / 0.9
Therefore, the volume of Al(OH)₃ needed for the reaction is 0.2 L
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