2 NaIO3(aq) + 5 NaHSO3(aq) → 3 NaHSO4(aq)
+ 2 Na2SO4(aq) + H2O(ℓ) + I2(aq)
What is the theoretical yield of I2 if you mixed 16.0 g of NaIO3 with 120 mL of 0.853 M NaHSO3?
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
mols NaIO3 = g/molar mass = ?
mols NaHSO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols NaIO3 to mols I2.
Do the same to convert mols NaHSO3 to mols I2.
It is likely these values will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that is the LR>
Using the smaller value, convert to grams I2. g = mols I2 x molar mass I2. This is the theoretical yield.
To find the theoretical yield of I2, we need to calculate how many moles of NaIO3 and NaHSO3 are present in the given amounts, and then determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by calculating the number of moles of NaIO3 using its molar mass.
Molar mass of NaIO3 = (22.99 g/mol × 2) + (126.9 g/mol) + (16.00 g/mol × 3)
= 45.98 g/mol + 126.9 g/mol + 48.00 g/mol
= 220.88 g/mol
Number of moles of NaIO3 = mass / molar mass
= 16.0 g / 220.88 g/mol
= 0.0724 mol
Next, we need to calculate the number of moles of NaHSO3 using its molarity and volume.
Molarity of NaHSO3 = 0.853 M
Volume of NaHSO3 = 120 mL = 0.120 L
Number of moles of NaHSO3 = Molarity × Volume
= 0.853 mol/L × 0.120 L
= 0.1024 mol
Now, we can use the stoichiometry of the balanced equation to determine the limiting reagent. The balanced equation shows that the ratio of NaIO3 to NaHSO3 is 2:5.
From the given amounts, there are 0.0724 moles of NaIO3 and 0.1024 moles of NaHSO3. To compare these values in the stoichiometric ratio, we can divide each value by the coefficient of the respective compound:
NaIO3: 0.0724 mol / 2 = 0.0362 mol
NaHSO3: 0.1024 mol / 5 = 0.0205 mol
Hence, we can see that NaHSO3 is the limiting reagent since there are fewer moles of NaHSO3 available compared to the stoichiometric ratio.
Now, using the stoichiometry of the balanced equation, we can determine the theoretical yield of I2.
From the balanced equation, we see that the ratio of NaIO3 to I2 is 2:1. Therefore, the number of moles of I2 that can be formed is half the number of moles of NaIO3.
Theoretical yield of I2 = 0.0362 mol (NaIO3) × 1/2
= 0.0181 mol
Finally, to convert the theoretical yield from moles to grams, we can use the molar mass of I2.
Molar mass of I2 = (126.9 g/mol × 2) = 253.8 g/mol
Theoretical yield of I2 (in grams) = 0.0181 mol × 253.8 g/mol
= 4.60 grams
Therefore, the theoretical yield of I2 in this reaction, if you mixed 16.0 g of NaIO3 with 120 mL of 0.853 M NaHSO3, is 4.60 grams.