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