Use Le Châtelier's principle to predict how the equilibrium for the weak base methylamine responds to the indicated changes. CH 3 NH 2 ( aq ) + H 2 O ( l ) − ⇀ ↽ − CH 3 NH + 3 ( aq ) + OH − ( aq ) Δ H ∘ rxn = 103 kJ / mol

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

c8h10 is your answerr