Given the following equations, determine the standard enthalpy of formation (ΔH∘f) for one mole of ICl3(g).I2(g)+3Cl2(g)⟶2ICl3(g)ΔH∘29 8=−214 kJI2(s)⟶I2(g)ΔH∘298=38 kJNote that iodine (I2) is a solid at standard conditions. Your answer should have two significant figures.

i just got this problem. it's 90.3

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

the answer is c i believe

dynamic equilibrium is the answer.