neutralization?
2016-05-31 1:47 am
In the neutralization of 3.00 M hydrochloric acid and sodium hydroxide, what volume, in mL, of the acid would you need to neutralize 50.0mL of 2.00 M sodium hydroxide.
回答 (1)
2016-05-31 1:54 am
HCl + NaOH → NaCl + H₂O
Mole ratio HCl : NaOH = 1 : 1
No. of moles of NaOH = (2.00 mol/L) × (50.0/1000 L) = 0.100 mol
No. of moles of HCl = (0.100 mol) × 1 = 0.100 mol
Volume of HCl used = (0.100 mol) / (3.00 mol/L) = 0.0333 L = 33.3 mL
Mole ratio HCl : NaOH = 1 : 1
No. of moles of NaOH = (2.00 mol/L) × (50.0/1000 L) = 0.100 mol
No. of moles of HCl = (0.100 mol) × 1 = 0.100 mol
Volume of HCl used = (0.100 mol) / (3.00 mol/L) = 0.0333 L = 33.3 mL
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