Question

The following process has been used to obtain iodine from oil-field brines in California.
$NaI + AgNO_3 \to AgI + NaNO_3 ....(1)$
$2AgI + Fe \to FeI_2 + 2Ag .... (2)$
$2FeI_2 + 3Cl_2 \to 2FeCl_3 + 2I_2 .... (3)$
How many grams of $AgNO_3$ are required in the first step for every $254 \,kg$ $I_2$ produced in the third step?

• A. $340 \times 10^3$
• B. $240 \times 10^3$
• C. $440 \times 10^3$
• D. $540 \times 10^3$

Answer 1

Answer: (A)

Balanced equation:
$NaI + Ag NO_3 \to AgI + NaNO_3\,....(1)$
$AgI + Fe \to FeI_2 + 2Ag\,....(2)$
$2FeI_2 + 3Cl_2 \to 2FeCl_2 + 2I_2 \,.....(3)$
From eq. $(3)$, we get
$\frac{\text{mole of} I_2}{2} = \frac{\text{mole of } FeI_2}{2}$
$\frac{\text{mole of} FeI_2}{1} = \frac{\text{mole of } AgI}{2}$
$\frac{\text{mole of} AgI}{1} = \frac{\text{mole of } AgNO_3}{1}$
$\therefore $ mole of $I_2 = $(mole of $FeI_2)$
$\left(\frac{\text{moles of} AgI}{2}\right) = \left( \frac{\text{mole of} Ag NO_{3}}{2} \right)$
$\frac{254\times10^{3}}{254} = \frac{\text{mole of} AgNO_{3}}{2} $
$ 2 \times10^{3} =$ mole of $AgNO_{3}$
$ = \frac{\text{mass of} AgNO_{3}}{\text{molar mass of} AgNO_{3}} $
Mass of $AgNO_{3 } = 170 \times\left(2 \times10^{3}\right)g $
$ = 340 \times10^{3} g$