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?
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.
Thanks! Sorry it's 125 ml of NaI
To determine the molarity of the resulting salt solution, we need to use the balanced chemical equation and the given information.
The balanced chemical equation for the reaction is:
2NaI (aq) + Cl2 → 2NaCl (aq) + I2 (s)
From the equation, we can see that 2 moles of NaI react with 1 mole of Cl2 to form 2 moles of NaCl. This means the stoichiometric ratio between NaI and NaCl is 2:2 or 1:1.
Given that the initial NaI solution has a concentration of 0.45 M, we need to calculate the moles of NaI present in the solution.
To do that, we need to multiply the volume of the NaI solution (in liters) by its concentration:
moles of NaI = volume (L) x concentration (M)
moles of NaI = 0.45 M x (volume in liters)
Since we don't have the volume stated in the question, we cannot calculate the exact molarity of the resulting salt solution. However, we can provide the general approach based on the stoichiometry.
Since the stoichiometric ratio between NaI and NaCl is 1:1, the moles of NaCl formed will be equal to the moles of NaI present initially.
So, the molarity of the resulting salt solution would be the same as the molarity of the initial NaI solution, which is 0.45 M.
Please note that this calculation assumes the reaction goes to completion and there are no other factors affecting the outcome. The actual molarity of the resulting salt solution may vary because it depends on the factors such as the extent of the reaction, temperature, and other conditions.