Question

How many liters of .100 m hcl hcl would be required to react with 5 grams of calcium hydroxide?

Answer 1

The  number  of liters  of  0.100 HCl that would  be required  to  react with  5g of Ca(OH)2 is calculated  as  follows

find the   moles of Ca(OH)2  used

= mass/molar mas
=  5g/ 74.09 g/mol  = 0.0675  moles
write the  equation  involved

that is Ca(OH)2 +2 HCl = CaCl2 + 2 H2O

by use of mole ratio  between  Ca(OH)2  to HCl  which  is 1:2  the  moles  of  HCl  is therefore = 2 x 0.0675 = 0.135  moles  of HCl

volume of HCl =  moles of HCL / molarity of HCl

= 0.135/ 0.100 = 1.35 L

Answer 2

the balanced equation for the reaction between HCl and Ca(OH)₂ is as follows;
Ca(OH)₂ + 2HCl ---> CaCl₂  + 2H₂O
stoichiometry of Ca(OH)₂ to HCl is 1:2
mass of Ca(OH)₂ reacting - 5 g
therefore number of moles of Ca(OH)₂ - 5 g / 74 g/mol = 0.068 mol 
according to molar ratio
number of HCl moles reacted = twice the number of Ca(OH)₂ reacted 
number of HCl moles required - 0.068 x 2 = 0.136 mol 
molarity if HCl solution - 0.100 M
there are 0.100 mol of HCl in 1 L 
therefore 0.136 mol in - 0.136 mol / 0.100 mol/L = 1.36 L
volume of 0.100 M HCl required - 1.36 L