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