Suppose you believe your unknown compound is naoh, and you prepare 20 ml of a 0.1 mol/l solution of it. the stockroom provides a 0.2 mol/l hcl solution as the titrant. what volume of hcl (in ml) is needed to reach the equivalence (stoichiometric) point if your unknown compound is naoh? naoh(aq) + hcl(aq) --> nacl(aq) + h2o(l)

Answer : Volume of HCl ( in ml)  is 10 ml.

Solution : Given,   Concentration of NaOH = 0.1 mol/L

                                Volume of NaOH = 20 ml

                                Concentration of HCl = 0.2 mol/L

                                 Volume of HCl =  ?

    Formula used :   Moles = Volume ×  Concentration

    In the reaction, we see that 1 mole of NaOH react with the 1 mole of HCl.

    so,

                       Moles of NaOH = Moles of HCl

    Moles of HCl = Moles of NaOH = Volume of NaOH ( in L) × Concentration of NaOH

    Converstion ml into L :      1000 ml = 1 L

             Moles of HCl =   × 0.1

                                    =  0.002 mol

             Volume of HCl = Moles / Comcentration of HCl

                                       =

                                       =  0.01 L

             Volume of HCl (in ml) =  0.01 × 1000 = 10 ml

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

a polar bond is formed when electrons are shared unevenly between atoms