In this redox reaction, identify the reducing agent. 2mgo → 2mg + o2 a. h2o b. mg c. mgo d. o

When you look up reducing agent, every definition you find uses the word 'substance.' For example, LiAlH4 is a powerful reducing agent, however only the H^- gets oxidized as LiAlH4 reduces some other element in a different compound.  

For your problem, let us write the equation this way:  

MgO +MgO ---> 2Mg + O2  

The O is oxidized from -2 to zero and is the specific atom that is the reducing agent, since it is the Mg that gets reduced from +2 to zero.  

However, should we pick MgO as the reducing agent (following the usual book definition) or should we pick the oxygen by itself?  

Based on the unusual nature of the problem, I will recommend you pick the O. However, be prepared for it to be the wrong choice and the correct answer to then be MgO.

Source(s):

ChemTeam

to summarize what the guy said, its d. O

this is right on plato, got 5/5

posting this for future users!

do u have any choices to choose from ? ?

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%