Ross attempted to titrate 30.0 mL of NaOH solution with 1.0 M HCl solution. The flask of NaOH solution contained 0.00300 mole of NaOH. Ross carelessly added 5.0 mL of HCl solution to the NaOH and observed blue litmus paper in the flask turn pink. What was the hydrogen ion concentration in the flask after all the HCl was added? A) 6.7 × 10-2 mol/L

B) 5.7 × 10-2 mol/L

C) 2.0 × 10-3 mol/L

D)1.0 × 10-7 mol/L

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

c (diamond),

sio2 (quartz)

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