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kenmyna
The  number of  Ml  of 0.45m HCl needed to neutralize  25 ml  of 1 ml KOH  is  calculated  as below

find the moles of KOH  used =  molarity  x volume

= 1  x25 = 25 moles
write the reacting  equation

KOH +HCL = KCl + H2O

by  use   of  mole ratio  between  KOH to HCl  which is 1:1 therefore the  moles of 25 moles

volume of HCl = moles/molarity

 25/0.45 = 55.6 ml  of HCL
the balanced equation for the reaction is as follows;
HCl + KOH ---> KCl + H₂O
stoichiometry of HCl to KOH is 1:1
number of KOH moles reacted - 1 mol/L x 0.025 L = 0.025 mol 
according to 1:1 molar ratio 
number of KOH moles reacted = number of HCl moles reacted 
therefore number of HCl moles reacted - 0.025 mol

molarity of HCl is 0.45 M
since 0.45 mol  are in 1000 mL 
therefore 0.025 mol are in a volume of 0.025 mol / 0.45 mol/L  = 55.5 mL 
55.6 mL of HCl is required to neutralise 

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