HCl has a molar concentration of 0.17m calculate the mass of aluminium hydroxide required to neutralise 50cm³ of the acid
The balanced equation for the neutralisation reaction between HCl and Al(OH)3 is:
3HCl + Al(OH)3 → AlCl3 + 3H2O
From the equation, we can see that 3 moles of HCl react with 1 mole of Al(OH)3. Therefore, the number of moles of HCl in 50 cm³ of 0.17 M HCl can be calculated as follows:
Number of moles of HCl = concentration × volume
= 0.17 mol/L × 0.05 L
= 0.0085 mol
Since 3 moles of HCl react with 1 mole of Al(OH)3, the number of moles of Al(OH)3 required can be calculated as follows:
Number of moles of Al(OH)3 = 0.0085 mol ÷ 3
= 0.00283 mol
The molar mass of Al(OH)3 is 78 g/mol. Therefore, the mass of Al(OH)3 required can be calculated as follows:
Mass of Al(OH)3 = number of moles × molar mass
= 0.00283 mol × 78 g/mol
= 0.22134 g (rounded to four significant figures)
Therefore, the mass of aluminium hydroxide required to neutralise 50 cm³ of 0.17 M HCl is approximately 0.2213 g.
To calculate the mass of aluminum hydroxide required to neutralize 50 cm³ of hydrochloric acid (HCl), we need to use the balanced chemical equation for the reaction between HCl and aluminum hydroxide (Al(OH)₃).
The balanced equation is as follows:
3HCl + Al(OH)₃ -> AlCl₃ + 3H₂O
From the equation, we can see that 3 moles of hydrochloric acid react with 1 mole of aluminum hydroxide.
First, we need to convert the given volume of HCl (50 cm³) to moles using the molar concentration of HCl.
Volume of HCl = 50 cm³
Molar concentration of HCl = 0.17 M
Molar concentration (M) is defined as moles of solute divided by volume of solution in liters. Therefore, we can calculate the number of moles of HCl as follows:
Moles of HCl = Molar concentration × Volume (in liters)
= 0.17 M × (50 cm³ / 1000 cm³/L)
= 0.0085 moles
Since the reaction ratio is 3 moles of HCl per 1 mole of aluminum hydroxide, we can determine the number of moles of aluminum hydroxide required using the mole ratio:
Moles of aluminum hydroxide = Moles of HCl / Reaction ratio
= 0.0085 moles / 3
= 0.00283 moles
The molar mass of aluminum hydroxide (Al(OH)₃) is:
Al = 26.98 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of Al(OH)₃ = (26.98 g/mol) + 3 × (16.00 g/mol) + 3 × (1.01 g/mol)
= 78.00 g/mol
Finally, we can calculate the mass of aluminum hydroxide needed using the moles of aluminum hydroxide and its molar mass:
Mass = Moles × Molar mass
= 0.00283 moles × 78.00 g/mol
≈ 0.22174 g
Therefore, approximately 0.222 grams of aluminum hydroxide is required to neutralize 50 cm³ of hydrochloric acid with a molar concentration of 0.17 M.