Which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?

K⁺ and NO₃⁻ are the spectator ions.

Explanation:

First, we will consider the molecular equation because is the easiest to balance.

K₂CrO₄(aq) + Ba(NO₃)₂(aq) → BaCrO₄(s) + 2 KNO₃(aq)

Then, we will write the full ionic equation, which includes all the ions and the molecular species.

2 K⁺(aq) + CrO₄²⁻(aq) + Ba²⁺(aq) + 2 NO₃⁻(aq) → BaCrO₄(s) + 2 K⁺(aq) + 2 NO₃⁻(aq)

Finally, we will write the net ionic equation, which includes only the ions that participate in the reaction and the molecular species. The missing ions are the spectator ones.

CrO₄²⁻(aq) + Ba²⁺(aq) → BaCrO₄(s)

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