For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Answer:
The amount of CaCl2 produced depends on the amount of HCl in the reaction.
Explanation:
The amount of HCl is used completelyin the reaction unlike CaCO3 which remains after reaction.
Answer is: <span>the molarity of HCl is </span>0.097 M.
Chemical reaction: LiOH + HCl → LiCl + H₂O.
V(HCl) = 13.60 mL - 1.25 mL = 12.35 mL.
V(LiOH) = 11.20 mL - 2.65 mL = 8.55 mL.
c(LiOH) = 0.140 M.
From chemical reaction: n(LiOH) : n(HCl) = 1 : 1.
c(HCl) · V(HCl) = c(LiOH) · V(LiOH).
c(HCl) = 8.55 mL · 0.140 M / 12.35 mL.
c(LiOH) = 0.097 M.
Explanation:
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