Given 24.3 g of LICI, Find moles of PbCI, produced. PbSO4 + LiCl → PbCl2 + Li2SO4
1 answer:
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Mass C₆H₈O₇ : 0.531484 g
<h3>Further explanation</h3>Reaction
3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)
MW NaHCO₃ : 84 g/mol
mass NaHCO₃ : 7.10² mg=0.7 g
mol NaHCO₃ :
mol C₆H₈O₇ :
MW C₆H₈O₇ : 192 g/mol
mass C₆H₈O₇ :
Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially = moles = 0.00600 moles
Neutralization reaction (back titration):
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration = moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.