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Smartphones contain roughly 25 milligrams of gold metal used in the circuitry of the device. Suppose the $Au(s)$ is plated by the following reduction reaction:

$${Au}_{2}{O}_{3}(s)+6{H}^{+}(aq)+6{e}^{-}\rightarrow 2Au(s)+3{H}_{2}O(l)$$

If the driving voltage can achieve a stable current of 70 milliamps, how long will the reaction need to be sustained in order to plate 25 milligrams of $Au(s)$?

$(1{ e}^{-}=1.602\times {10}^{-19}\text{ coulombs},1\text{ ampere}=\frac{1\text{ coulomb}}{1\text{ second}},1\text{ mole}=6.022\times {10}^{23})$


$1.1\times {10}^{2}\text{ seconds}$


$5.2\times {10}^{2}\text{ seconds}$


$1.7\times {10}^{2}\text{ seconds}$


$8.0\times {10}^{2}\text{ seconds}$

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