Answer: Reducing agent in the given reaction is 
.
Explanation:
A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.
In the given reaction, oxidation state of sulfur in is +2 and 
has 0 oxidation state.
In 
oxidation state of S is 2.5 and in 
oxidation state of I is -1.
Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.
Thus, we can conclude that reducing agent in the given reaction is .
<span>Start with the number of grams of each element, given in the problem.Convert the mass of each element to moles using the molar mass from the periodic table.Divide each mole value by the smallest number of moles calculated.<span>Round to the nearest whole number.</span></span>
Find it on google i’m pretty sure i saw it somewhere so sorry this doesn’t help
First off chlorine is not a metal so you can ignore that one.
Sodium and Rubidium are in group 1 of the periodic table and Magnesium is in group 2.
Group one metals are more reactive than group two because it is harder for the group two metals to lose their 2 valence (outer most) electrons.
As you go down group 1 there is an increase in the reactivity this is because as you go down there is an increase in the atomic radius which leads to more shielding. This weakens the electrostatic forces of attraction making it easier to lose the outermost electrons, therefore they are more reactive.
