You didn't finish the problem. How many mL of the NaI?
moles NaI = M x L
moles NaI = moles NaCl.
M NaCl = mole/L soln.
2NaI (aq) + Cl2 = 2NaCl (aq) + I2 (s)
If the ml of 0.45 M NaI are reacted with chlorine gas:
What will be the molarity of the resulting salt solution?
2 answers
Thanks! Sorry it's 125 ml of NaI