Calculate the volume of 2.5M Mg(OH)2 necessary to neutralize 34mL of .75M HCl?
回答 (1)
2016-05-06 6:04 pm
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
Mole ratio Mg(OH)₂ : HCl = 1 : 2
No. of milli-moles of HCl = (0.75 mmol/mL) × (34 mL) = 25.5 mmol
No. of milli-moles of Mg(OH)₂ = (25.5 mmol) × (1/2) = 25.5/2 mmol
Volume of Mg(OH)₂ = (25.5/2 mmol) / (2.5 mmol/mL) = 5.1 mL
Mole ratio Mg(OH)₂ : HCl = 1 : 2
No. of milli-moles of HCl = (0.75 mmol/mL) × (34 mL) = 25.5 mmol
No. of milli-moles of Mg(OH)₂ = (25.5 mmol) × (1/2) = 25.5/2 mmol
Volume of Mg(OH)₂ = (25.5/2 mmol) / (2.5 mmol/mL) = 5.1 mL
收錄日期: 2021-04-18 14:47:57
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