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Calculate the mass (in grams) of $\mathrm{AgNO}_3$ needed in step I of the given process for every 127.5 Kg $\mathrm{I_2}$ formed in step III.

$\mathrm{NaI}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgI}+\mathrm{NaNO}_3 \quad ............(1)$

$2 \mathrm{AgI}+\mathrm{Fe} \rightarrow \mathrm{FeI}{ }_2+2 \mathrm{Ag} \quad ............(2)$

$\mathrm{2 \mathrm{FeI}_2+3 \mathrm{Cl}_2 \rightarrow 2 \mathrm{FeCl}_3+2 I_2} \quad ............(3)$

Option: 1
$170\times10^{4}$

Option: 2
$170 \times 10^3$

Option: 3
$17 \times 10^3$

Option: 4
$17 \times 10^5$

Answers (1)

best_answer

On multiplying eq 1 by 4 & eq 2 by 2 and adding all the equations.

$\begin{aligned} 4 \mathrm{NaI}+4 \mathrm{AgNO}_3+2 \mathrm{Fe}+3 \mathrm{Cl}_2 \rightarrow 4 \mathrm{NaNO}_3+4 \mathrm{Ag} & +2 \mathrm{FeCl_{3}} +2 \mathrm{I}_2 \\ \end{aligned}$

As per the equation,

Moles of

$\begin{aligned} \text{I}_{2}& =\frac{127.5 \times 10^3}{254} \\ \\& =0.5 \times 10^3 \\ \\& =5 \times 10^2 \end{aligned}$

To produce 2 moles of $\mathrm{I_{2}}\rightarrow 4$ moles of $\mathrm{AgNO_{3}}$ is used

For $5 \times 10^{2}$ moles $\mathrm{I_2=\frac{4}{2} \times 5 \times 10^2}$

$10^{3}$ mole $\mathrm{AgNO_{3}}$ is required.

Required mass of $\mathrm{AgNO_{3}}$ $\mathrm{=n \times M_{AgNO_{3}}}$

$=10^{3}\times 170$

Posted by

Rishi

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