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.
HCl has a molar concentration of 0.17m calculate the mass of aluminium hydroxide required to neutralise 50cm³ of the acid
1 answer
1 answer