Consider the reaction: hcl(aq) + naoh(aq) → naoh(aq) + h2o(l) how much 0.113 m naoh solution will completely neutralize 1.25 l of 0.228 m hcl solution?

The  much of 0.113 M NaOH  solution that  is required  to  neutralize  is  2.522 L of NaOH

    calculation

HCl (aq) + NaOH (aq) → NaOH (aq)  + H2O  (l)

Step 1:  calculate the   moles of  HCl

  moles= molarity  x volume in liters

volume in liters =  1.25 L

molarity =  0.228 M  =  0.228 mol/ L

moles is therefore = 0.228 mol/L  x 1.25 L = 0.285  mols

Step  2:  use the mole ratio to  calculate  the   moles of NaOH

The  mole ratio   of  HCl : NaOH  is 1:1  therefore  the moles  of NaOH= 0.285 moles

Step  3:  calculate the volume  of  NaOH

volume=moles/molarity

molarity  0.113 M  = 0.113 mol/l

= 0.285 moles /0.113 Mol/  L = 2.522 L  of NaOH

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year