<u>Given:</u>
Mass of Na = 115 g
Excess Cl2
<u>To determine:</u>
Mass of NaCl produced
<u>Explanation:</u>
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