Write out the complete and balanced equation of the reaction in order to find the ratio of moles required
#H_2SO_4(aq) + Ca(OH)_2(aq) → CaSO_4(s)+2H_2O(l)#
State symbols in the equation above
#s# - Solid
#l# - Liquid
#aq# - Aqueous (dissolved in water)
Relative Formula Mass of Calcium Hydroxide #(Ca(OH)_2)#
#Ca# = #40#
#O# = #16#
#H#= #1#
#40#x#1# = #40#
#16#x#2# = #32#
#1#x#2# = #2#
#40+32+2 = 74#
#RFM# = #74gmol^-1#
To find the number of moles of Calcium Hydroxide #(Ca(OH)_2)# when we have the mass and relative formula mass we use the formula and substitute in the values
#n = m/M#
Where
#n# = number of moles (#mol#)
#m# = mass (#grams#)
#M# = relative formula mass (#gmol^-1#)
#n# = #2.5 / 74#
Number of moles in (#1#) Calcium Hydroxide = #5/148#
Since the reactants are in a #1:1# Ratio we can state that Sulfuric Acid #(H_2SO_4)# and Calcium Hydroxide #(Ca(OH)_2)# have the same number of moles.
Number of moles in (#1#) Calcium Hydroxide #(Ca(OH)_2)# and (#1#) Sulfuric Acid
#∴# = #5/148#
Now that we have the number of moles and the concentration of Sulfuric Acid #(H_2SO_4)# we can find the volume.
To do this use the formula
#n# = #c# x #v/1000#
Where
#n# = number of moles (#mol#)
#c# = concentration (#moldm^-3#)
#v# = volume (#cm^3#)
Substitute in the values and rearrange for # v#
#5/148##mol# = #0.125##moldm^-3*v/1000#
#5/148mol#x#1000# = #0.125##moldm^-3*v#
#1250/37mol# = #0.125##moldm^-3\8v#
#(1250/37mol)/( 0.125moldm^-3) = v#
#v# = #270.27# #cm^3#
To mL
#v# = #270.27##mL#
#:.# the volume of #0.125M# #H_2SO_4# required to neutralize #2.5g# of #Ca(OH)_2# is #270.27mL#
