To determine the mole ratio of Na2TeO3, NaI, and HCl in the oxidation reaction of iodide ions (I-) to iodine (I2) using sodium tellurite (Na2TeO3) in the presence of hydrochloric acid (HCl), we first need to write the balanced chemical equation for the reaction. This will help us identify the stoichiometric coefficients, which represent the mole ratios of the reactants and products involved.
Step-by-Step Balancing of the Reaction
The overall reaction can be represented as follows:
Na2TeO3 + HCl + I- → I2 + Te + NaCl + H2O
Now, let’s break down the balancing process:
Identifying Reactants and Products
- Reactants: Na2TeO3, HCl, I-
- Products: I2, Te, NaCl, H2O
Balancing the Equation
1. **Balance Iodine (I):** The left side has I- ions, and we need to produce I2. To balance iodine, we need 2 I- ions for every I2 produced.
2. **Balance Sodium (Na):** There are 2 sodium atoms in Na2TeO3, so we need 2 NaCl on the product side.
3. **Balance Tellurium (Te):** There is 1 tellurium atom in Na2TeO3, which will yield 1 Te in the products.
4. **Balance Chlorine (Cl):** Since we have 2 HCl contributing to the 2 NaCl, we will need 2 HCl on the reactant side.
5. **Balance Hydrogen (H) and Oxygen (O):** Finally, we need to ensure that the hydrogen and oxygen atoms are balanced. The 2 HCl will provide 2 H atoms, which can combine with oxygen to form H2O.
After balancing, the equation looks like this:
Na2TeO3 + 2 HCl + 2 I- → I2 + Te + 2 NaCl + H2O
Determining the Mole Ratios
From the balanced equation, we can now extract the mole ratios:
- Na2TeO3: 1 mole
- NaI: 2 moles (since we have 2 I- ions)
- HCl: 2 moles
Thus, the mole ratio of Na2TeO3:NaI:HCl is:
1:2:2
Conclusion
In summary, the balanced equation reveals that for every mole of sodium tellurite, we require two moles of iodide ions and two moles of hydrochloric acid. This stoichiometric relationship is crucial for understanding how to scale the reaction for practical applications, such as in laboratory settings or industrial processes.