K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s) net ionic equation

SO4 (aq) + Sr (aq) >> SrSO4 (s)

Explanation:

K+1 SO4-2 + Sr +2 I2- >> sK +1

(you dont separate and write down the ions for Sr SO4 because its a solid)

Cancel out the same ions from both sides and write down what your left with.

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