How many milliliters of 0.100 M HClO₃ are required to neutralize 40.0 mL of 0.140 M KOH?
Balanced chemical equation is:
KOH + HClO3 ---> KClO3 + H2O
Here:
M(KOH)=0.14 M
M(HClO3)=0.1 M
V(KOH)=40.0 mL
According to balanced reaction:
1*number of mol of KOH =1*number of mol of HClO3
1*M(KOH)*V(KOH) =1*M(HClO3)*V(HClO3)
1*0.14 M *40.0 mL = 1*0.1M *V(HClO3)
V(HClO3) = 56.0 mL
Answer: 56.0 mL
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