What volume of this freshly prepared 0.500 M HNO3 would be needed to neutralize 2.00 g of Ca(OH)2?
回答 (1)
2016-07-19 3:23 pm
2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
OR: Mole ratio HNO₃ : Ca(OH)₂ = 2 : 1
No. of moles of Ca(OH)₂ = (2.00 g) / (74.09 g/mol) = 2.00/74.09 mol
No. of moles of HNO₃ = (2.00/74.09 mol) × 2 = 4.00/74.09 mol
Volume of HNO₃ = (4.00/74.09 mol) / (0.500 mol/L) = 0.108 L = 108 mL
OR: Mole ratio HNO₃ : Ca(OH)₂ = 2 : 1
No. of moles of Ca(OH)₂ = (2.00 g) / (74.09 g/mol) = 2.00/74.09 mol
No. of moles of HNO₃ = (2.00/74.09 mol) × 2 = 4.00/74.09 mol
Volume of HNO₃ = (4.00/74.09 mol) / (0.500 mol/L) = 0.108 L = 108 mL
收錄日期: 2021-04-18 15:21:45
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