Answer:
15L of 0.40M NaCl(aq) solution
Explanation:
2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g)
2Na⁺(aq) + 2Clˉ(aq) + 2H₂O(l) → 2Na⁺(aq) + 2OHˉ(aq) + Cl₂(g)
Na⁺(aq) is a spectator ion in the given reaction and does not enter into the reaction process…
Net Ionic Equation is then 2Clˉ(aq) + 2H₂O(l) → 2OHˉ(aq) + Cl₂(g)
From Rxn, 2 moles Clˉ(aq) is needed to produce 1 mole of Cl₂(g)
Therefore, 6 moles Clˉ(aq) is needed to produce 3 moles of Cl₂(g)
That is, 6 moles NaCl(aq) → 6 moles Clˉ(aq) = 0.40M x V(NaCl)liters
V(NaCl) liters = 6 moles Clˉ(aq)/(0.40mole/liter) = 15 liters of 0.40M NaCl(aq)
H20-h&o
Gold-au
Nitric acid-h,n,o
Cobalt-co
Oxygen-o
MAKE SURE THEY ARE CAPITALIZED