The grams of Al(OH)₃ can be neutralized by 300ml of 0.250m hydrochloric acid is 1.9500 gram
The moles of HCl = Molarity × volume in liters
= 0.250 × 0.3
moles of Hcl = 0.075 mol
The balanced equation may be
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
1 mol of Al(OH)₃ neutralizes 3 mol of Hcl
To neutralizes 0.075 mol of Hcl
= 0.075 / 3
= 0.025 mol of Al(OH)₃
Mass of Al(OH)₃ = Mole × molar mass
= 0.025 × 78.0025
= 1.9500 gram
Hence, the gram of Al(OH)₃ can be neutralized by 300 ml of 0.250 HCl is 1.9500 gram.
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