If 390 mL of 0.0410 M aqueous NaI(aq) and 1.36 g of liquid Br2 are reacted stoichiometrically according to the equation, what mass (g) of NaI(aq) remained?

2 NaI(aq) + Br2(l) �¨ 2 NaBr(aq) + I2(s)
Molar Mass (g)
NaI(aq) 149.89
Br2 159.81


Molar Volume (L)
22.400 at STP
Gas Constant
0.0821

1. Convert 390 mL of 0.0410 M NaI to moles NaI. M x L = ?? moles. I found 0.01599 moles NaI.

2. Convert 1.36 g Br2 to moles. moles = g/molar mass. I found 0.008510 moles Br2.
3. Which is the limiting reagent?
a. moles NaBr formed if all NaI is used is
0.01599 x (2 mols NaBr/2 moles NaI) = 0.01599 x 1/1 = 0.01599 moles NaBr.

b. moles NaBr formed if all of the Br2 is used is
0.008510 x (2 moles NaBr/1 mol Br2)= 0.008510 x 2 = 0.01702 moles NaBr.

c. The smaller number is the one to use so NaI is the limiting reagent which means that all of it used and none remains.
Check my numbers. Check my thinking.

To solve this problem, we can use the concept of stoichiometry and the given molarities and masses.

1. Convert the given volume of NaI(aq) to moles:
Given:
- Volume of NaI(aq) = 390 mL = 0.390 L
- Molarity of NaI(aq) = 0.0410 M

To convert the volume to moles, use the formula: moles = volume (L) × molarity.
moles of NaI = 0.390 L × 0.0410 M = 0.01599 moles

2. Convert the given mass of Br2 to moles:
Given:
- Mass of Br2 = 1.36 g
- Molar mass of Br2 = 159.81 g/mol

To convert the mass to moles, use the formula: moles = mass (g) ÷ molar mass.
moles of Br2 = 1.36 g ÷ 159.81 g/mol = 0.00851 moles

3. Determine the stoichiometry of the reaction:
According to the balanced equation, 2 moles of NaI react with 1 mole of Br2 to form 2 moles of NaBr and 1 mole of I2.

4. Calculate the limiting reagent:
The limiting reagent is the one that is completely consumed in the reaction and determines the amount of product formed. We compare the moles of NaI and Br2 to find the limiting reagent.

From step 1: moles of NaI = 0.01599 moles
From step 2: moles of Br2 = 0.00851 moles

Since the stoichiometric ratio is 2 NaI: 1 Br2, we divide the moles of Br2 by 2.
Adjusted moles of Br2 = 0.00851 moles ÷ 2 = 0.004255 moles

Comparing the moles of NaI and the adjusted moles of Br2, it is clear that NaI is the limiting reagent.

5. Calculate the moles of NaI used in the reaction:
Since NaI is the limiting reagent, all of the moles of NaI will be consumed in the reaction. Therefore, the moles of NaI used in the reaction equal the initial moles of NaI.

Moles of NaI used = 0.01599 moles

6. Convert the moles of NaI used to grams:
Given:
- Molar mass of NaI(aq) = 149.89 g/mol

To convert moles to grams, use the formula: mass (g) = moles × molar mass.
Mass of NaI used = 0.01599 moles × 149.89 g/mol = 2.397 g

7. Calculate the mass of NaI remaining:
To find the mass of NaI remaining, subtract the mass of NaI used from the initial mass of NaI.

Given:
- Initial mass of NaI(aq) = ?

Since we are not given the initial mass of NaI(aq), we cannot determine the mass of NaI remaining.

In conclusion, without knowing the initial mass of NaI(aq), we cannot calculate the mass of NaI remaining.