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.
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?
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1 answer