Identify the equations that show ionization. Check all that apply.
A) NaCl(s) → Na+ (aq)+Cl(aq)

B) HC (9)+H2O(l) → H30+ (aq) + Cr (aq)

C) O2(g) → O2(aq)

D) CO2(g) + 2H2O(l) → HCO3(aq) +H30+ (aq)

E)CaCl2(s) → Ca2+ (aq) + 2Cr (aq)

B. HCl(g)+H2O(I)—>H3O+(aq)+Cl-(aq)

D. CO2(g)+2H2O(I)—>HCO3-(aq)+H3O+(aq)

Explanation: on edge

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g