If 0.870 mol of liquid Br2 and 720 mL of 0.958 M aqueous NaI are reacted stoichiometrically according to the balanced equation, how many grams of solid I2 are produced?
2NaI(aq) + Br2(l) → 2NaBr(aq) + I2(s)
help please. :)
A limiting reagent problem.
1. Write the balanced equation.
2. You have Br already in mols.
3. Convert NaI to moles. moles = M x L.
4a. Using the coefficients in the balanced equation, convert moles Br2 to moles I2.
4b. Using the same procedure, convert moles NaI to moles I2.
4c. It is likely that the moles I2 produced will not be the same from 4a and 4b. Obviously, one of them in not correct. The correct answer in limiting reagent problems is ALWAYS the smaller value.
5. Now convert moles I2 from 4c to grams. g = moles x molar mass.
To determine the number of grams of solid I2 produced, we need to follow these steps:
Step 1: Calculate the number of moles of NaI available in the solution.
Given that the volume is 720 mL (0.720 L) and the concentration is 0.958 M, we can calculate the number of moles of NaI using the formula:
Moles of NaI = concentration (M) × volume (L)
Moles of NaI = 0.958 M × 0.720 L = 0.6906 moles
Step 2: Determine the limiting reagent.
To find the limiting reagent, we need to compare the molar ratios between Br2 and NaI in the balanced equation. The balanced equation tells us that the ratio is 1:2 (1 mole of Br2 reacts with 2 moles of NaI).
So, we need to find how many moles of Br2 are required:
Moles of Br2 = 2 × Moles of NaI = 2 × 0.6906 moles = 1.3812 moles
Given that we have 0.870 moles of Br2, it is clear that Br2 is the limiting reagent because we have less than the required amount.
Step 3: Calculate the number of moles of I2 produced.
Since we know from the balanced equation that 1 mole of Br2 produces 1 mole of I2, we can conclude that 0.870 moles of Br2 will produce 0.870 moles of I2.
Step 4: Convert moles of I2 to grams.
To convert moles of I2 to grams, we need to use the molar mass of iodine (I2). The molar mass of I2 is approximately 253.8089 g/mol.
Grams of I2 = Moles of I2 × Molar mass of I2
Grams of I2 = 0.870 moles × 253.8089 g/mol = 220.843 g
Therefore, approximately 220.843 grams of solid I2 are produced in this reaction.