For 2Na (s) + Cl2 (g) - 2NaCI (s), how many grams of sodium Chloride will be made from 115 g of sodium metal when reacted with excess chlorine gas?

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