Question

NaIO3 + 6 HI ------ 3 I2 + NaI + 3 H2OCalculate the number of moles of iodine that can be made from 16.4 grams of NaIO3

Answer 1

The number of moles of iodine is 0.249 mol

Step-by-step explanation

Given that;

The mass of sodium iodate is 16.4 grams

Follows the steps below to find the number of moles of iodine

Step 1; Write a balanced equation of reaction

`\text{ NaIO}_{3\text{ }}\text{ + 6HI }\rightarrow\text{ 3I}_2\text{ + NaI + 3H}_2O`

Step 2; Find the moles of NaIO3 using the below formula

`\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of NaOl3 is 197.89 g/mol

`\begin{gathered} \text{ mole = }\frac{\text{ 16.4}}{\text{ 197.89}} \\ \text{ mole = 0.083 mol} \end{gathered}`

Step 3; Find the number of moles of iodine using stoichiometry ratio

In the equation above, 1 mole NaIO3 reacts to give 3 moles I2

Let x represents the number of moles of iodine

`\begin{gathered} \text{ 1 mole NaIO}_3\text{ }\rightarrow\text{ 3 moles I}_2 \\ \text{ 0.083 mol NaIO}_3\text{ }\rightarrow\text{ x mol I}_2 \\ \text{ cross multiply} \\ \text{ 1 mole NaIO}_3\text{ }*\text{ x mole I}_2\text{ = 3 moles I}_2\text{ }*\text{ 0.083 mol NaIO}_(3e) \\ \text{ Isolate x mole I}_2 \\ \text{ x mole I}_2\text{ = }\frac{3\text{ moles I}_2\text{ }*0.083mol\cancel{NaIO_3}}{1mole\cancel{NaIO_3}} \\ \\ \text{ x mole I}_2\text{ = 3 }*\text{ 0.083} \\ \text{ x mole I}_2\text{ = 0.249 mol} \end{gathered}`

Therefore, the number of moles of iodine is 0.249 mol