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

Explanation:

To calculate the moles :

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

$\text{Moles of} I_2=\frac{55.6g}{254g/mol}=0.219moles$

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

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 =$\frac{2}{1}\times 0.219=0.438moles$ of $NaI$

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

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