#(i)# What volume of #5.00*mol*L^-1# #HCl(aq)# is required to react with a #17.0*g# mass of sodium hydroxide. #(ii)# what volume of sodium sulfide at #0.100*mol*L^-1#, is required to precipitate a #33.0*mL# volume of #0.100*mol*L^-1# of #AgNO_3#?

Our basic equation is......

#"Concentration"="Moles of solute"/"Volume of solution"#.

And thus by taking a product or a quotient, we can get the number of moles or the volume of solution.

#NaOH(s) + HCl(aq) rarr NaCl(aq) + H_2O(l)#

And thus, for the first equation there is 1:1 stoichiometry. And moles of #NaOH# are equivalent to moles of #HCl#.

#"Moles of NaOH"-=(17.0*g)/(40.0*g*mol^-1)=0.425*mol#

And by the give stoichiometry, we need ONE equiv of #HCl(aq)#.

We are given #5.00*mol*L^-1# #HCl(aq)#, and thus we solve the quotient......

#(0.425*cancel(mol))/(5.00*cancel(mol*L^-1))xx10^3*mL*cancel(L^-1)=85.0*mL#.

Clear?

For the second scenario, you must simply KNOW that #Ag_2S# (like most sulfides save those of the alkali metals) is INSOLUBLE.

#2AgNO_3(aq) + Na_2S(aq) rarr Ag_2S(s)darr+2NaNO_3(aq)#

The net ionic equation is simply......

#2Ag^+ + S^(2-) rarr Ag_2S(s)darr#

#"Moles of"# #AgNO_3=33.0xx10^-3*Lxx0.100*mol*L^-1=#
#3.30xx10^-3*mol#

And thus we need half an equiv of #Na_2S#, i.e.

#(1/2xx3.3xx10^-3*mol)/(0.100*mol*L^-1)xx10^3*mL*L^-1#

#"under 17 mL"# of the sodium sulfide solution.........