“10.00mL of a sample of concentrated hydrochloride acid is diluted to 100.00mL. If a 20.00mL portion required 15.23mL of 1.5 M NaOH to reach the endpoint, what is the concentration of the original, undiluted HCl?” I thought I knew what I was doing here but I ended with 114 M HCl. Can anyone point me in the right direction?

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

um i think it might be a car?

answermeridional ;