How many milliliters of 0.183 m hcl would be required to titrate 5.93 g koh?

  Molar mass:
KOH = 56.0 g/mol

Number of moles of KOH :

5.93  / 36.5 => 0.1624 moles

KOH + HCl = KCl + H₂O 

1 mole KOH > 1 mole  HCl
0.1624 moles KOH > ?

moles HCl = 0.1624 x 1 / 1

= 0.1624 moles of HCl

V ( HCl ) = moles / molarity

V(HCl) = 0.1624 / 0.183

V (HCl) = 0.887 L x 1000 = 887 mL

hope this helps!

The answer to that would be magnesium