I been trying to look at examples of similar question like this but I haven't found one sadly, how I do this one?
An antacid tablet, such as Amphojel, may be taken to reduce excess stomach acid, which is a 0.28 M HCl solution. If one dose of Amphojel contains 410 mg of Al(OH)3, what volume, in milliliters, of stomach acid will be neutralized?3HCl(aq)+Al(OH)3(s)→3H2O(l)+AlCl3(aq)
How I do this? Chem help please?
2020-05-15 2:05 pm
回答 (2)
2020-05-15 2:16 pm
Molar mass of Al(OH)₃ = 27.0 + 1.0×3 + 16.0×3 = 78.0 g/mol
Millimoles of Al(OH)₃ = (410 mg) / (78.0 g/mol) = 5.256 mmol
3HCl(aq) + Al(OH)₃(s) → 3H₂O(ℓ) + AlCl3(aq)
Mole ratio HCl : Al(OH)₃ = 3 : 1
Millimoles of HCl consumed = (5.256 mmol) × 3 = 15.77 mmol
Volume of HCl neutralized = (15.77 mmol) / (0.28 mmol/mL) = 56.3 mL
====
OR:
(410/1000 g Al(OH)₃) × (1 mol Al(OH)₃ / 78.0 g Al(OH)₃) × (3 mol HCl / 1 mol Al(OH)₃) × (1000 mL stomach acid / 0.28 mol HCl)
= 56.3 mL stomach acid
Millimoles of Al(OH)₃ = (410 mg) / (78.0 g/mol) = 5.256 mmol
3HCl(aq) + Al(OH)₃(s) → 3H₂O(ℓ) + AlCl3(aq)
Mole ratio HCl : Al(OH)₃ = 3 : 1
Millimoles of HCl consumed = (5.256 mmol) × 3 = 15.77 mmol
Volume of HCl neutralized = (15.77 mmol) / (0.28 mmol/mL) = 56.3 mL
====
OR:
(410/1000 g Al(OH)₃) × (1 mol Al(OH)₃ / 78.0 g Al(OH)₃) × (3 mol HCl / 1 mol Al(OH)₃) × (1000 mL stomach acid / 0.28 mol HCl)
= 56.3 mL stomach acid
2020-05-15 2:32 pm
Given the mass, start there and multiply by (1)
410 mg of Al(OH)3 (1g/1000mg) (1 mol/78 g) (1L/0.28 mol) (1000mL/1L) (3 mol HCL/1 mol Al(OH)3) = 56.31868131 mL
Now round to 2 sigfigs>>> 56. mL
410 mg of Al(OH)3 (1g/1000mg) (1 mol/78 g) (1L/0.28 mol) (1000mL/1L) (3 mol HCL/1 mol Al(OH)3) = 56.31868131 mL
Now round to 2 sigfigs>>> 56. mL
收錄日期: 2021-05-01 09:39:09
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