Question

Iodine is prepared both in the laboratory and commercially by adding Cl2(g) to an aqueous solution containing sodium iodide: 2NaI(aq) +Cl2(g) → I2(s) +2NaCl(aq) How many grams of iodide, NaI, must be used to produce 55.6 g of iodine, I2?

Answer 1

Answer: 65.7 gram

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} I_2=(55.6g)/(254g/mol)=0.219moles`

`2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)`

According to stoichiometry :

1 mole of
`I_2` are produced by = 2 moles of
`NaI`

Thus 0.219 moles of
`I_2` will be produced by =
`(2)/(1)* 0.219=0.438moles` of
`NaI`

Mass of
`NaI=moles* {\text {Molar mass}}=0.438moles* 150g/mol=65.7g`

Thus 65.7 g of NaI, must be used to produce 55.6 g of iodine