Question

What is the limiting reagent in a reaction of 36g of sodium Chloride and 58g of Potassium Iodide?

Answer 1

Sol:-

Given,

Mass of

Sodium chloride (NaCl) = 36g

Potassium iodide (KI) = 58g

molar mass of

NaCl = (23.0 + 35.5)g/mol = 58.5 g/mol

KI = (39.1 + 126.9) g/mol = 166 g/mol

So,

No. of moles of

NaCl = 36g/(58.5g/mol) = 0.6154 mol

KI = 58g/(166g/mol) = 0.3494 mol

as per stoichiometry, 1 mol NaCl react with 1 mol KI

(NaCl + KI NaI + KCl)

So,

0.3434 mol KI react with 0.3494 mol NaCl.

Hence, there will be excess NaCl (0.6154 mol - 0.3494 mol = 0.266 mol) as moles of NaCl is more than moles of KI. And KI will completely react with NaCl.

So,

Limiting reagent is KI.