Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
Answer:
95.7 g CO to the nearest tenth.
Explanation:
2C + O2 ---> 2CO
Using relative atomic masses:
24 g C produces 2*12 + 2*16 g CO.
So 41 g produces ( (2*12 + 2*16) * 41 ) / 24
= 95.7 g CO,
<span>Good Morning!
</span><span>reactants
</span><span>
Substances that make up a chemical reaction, in order to be combined or separated, are called reagents. These reagents can be of the most varied types and origins. The result of this reaction is called a "product."
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