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.
Pv=nRT
where,p=199, R(constant)=8.314, V=4.67 T=30C=293K
n=pv/RT=0.38 moles
Cost per mole
Table salt : Rs 0.878
Table sugar : Rs 23.63
<h3>Further explanation</h3>
Given
Cost table salt (NaCl) = 15/kg
Cost table sugar(sucrose-C12H22O11) = 69/kg
Required
cost per mole
Solution
mol of 1 kg Table salt(NaCl ,MW= 58.5 g/mol) :
mol of 1 kg Table sugar(C12H22O11 ,MW= 342 g/mol) :
