21.65 mL of 0.0237M Na2CO3 was the average titre to neutralise a HCl solution ( ie to find the end-point of the reaction)
The reaction is: Na2CO3 + 2HCl → 2 NaCl + H2CO3
Calculate the number of moles (n) of HCl reacted (read below)?
2016-06-19 4:17 pm
回答 (1)
2016-06-19 5:05 pm
Na₂CO₃ + 2HCl → 2NaCl + H₂CO₃
OR: Mole ratio Na₂CO₃ : HCl = 1 : 2
No. of moles of Na₂CO₃ reacted = (0.0237 mol/L) × (21.65/1000 L)
No. of moles of HCl reacted = (0.0237 mol/L) × (21.65/1000 L) × 2 = 0.00103 mol
OR: Mole ratio Na₂CO₃ : HCl = 1 : 2
No. of moles of Na₂CO₃ reacted = (0.0237 mol/L) × (21.65/1000 L)
No. of moles of HCl reacted = (0.0237 mol/L) × (21.65/1000 L) × 2 = 0.00103 mol
收錄日期: 2021-04-18 15:06:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160619081722AA4pMRr