A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1876 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 15.95 mL of the standard solution. What is the concentration of the original sulfuric acid solution?

hope you got the answer

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

the answer your looking for is c).