Question

The following process has been used to obtain iodine from oil-field brines in California.
$NaI+AgN{O}_3\to AgI+NaN{O}_3....(1)$
$2AgI+Fe\to Fe{I}_2+2Ag....(2)$
$2Fe{I}_2+3C{l}_2\to 2FeC{l}_3+2I_2....(3)$
How many grams of $AgN{O}_3$ are required in the first step for every $254kg$ $I_2$ 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:
$NaI+AgN{O}_3\to AgI+NaN{O}_3....(1)$
$AgI+Fe\to Fe{I}_2+2Ag....(2)$
$2Fe{I}_2+3C{l}_2\to 2FeC{l}_2+2I_2.....(3)$
From eq. $(3)$, we get
$\frac{\text{mole of}{I}_2}{2}=\frac{\text{mole of }Fe{I}_2}{2}$
$\frac{\text{mole of}Fe{I}_2}{1}=\frac{\text{mole of }AgI}{2}$
$\frac{\text{mole of}AgI}{1}=\frac{\text{mole of }AgN{O}_3}{1}$
$\therefore$ mole of $I_2=$(mole of $Fe{I}_2)$
$\left(\frac{\text{moles of}AgI}{2}\right)=\left(\frac{\text{mole of}AgN{O}_3}{2}\right)$
$\frac{254\times 10^3}{254}=\frac{\text{mole of}AgN{O}_3}{2}$
$2\times 10^3=$ mole of $AgN{O}_3$
$=\frac{\text{mass of}AgN{O}_3}{\text{molar mass of}AgN{O}_3}$
Mass of $AgN{O}_3=170\times \left(2\times 10^3\right)g$
$=340\times 10^{3}g$