MolSynapsequestion bank

MolSynapse / Question bank / Mole Concept / Limiting reagent, percent yield and percent excess

JEE Advanced levelSingle correct optionDifficulty 3 of 3mc068

$11.5 g$ of $Na$ reacts with $21.3 g$ of $Cl X_{2}$ in $2 Na + Cl X_{2} 2 NaCl$ ( $Na = 23$ , $Cl = 35.5$ ). The percent excess of the reagent in excess (relative to its stoichiometric requirement) and the mass of $NaCl$ formed are

A $16.7%$ and $29.25 g$
B $20%$ and $35.10 g$
C $20%$ and $29.25 g$
D $40%$ and $14.62 g$

Answer: C

Na

= 0.5 mol;

Cl X_{2}

= 0.3 mol.

Na

needs 0.25 mol

Cl X_{2}

, so

Na

limits and

Cl X_{2}

is in excess. Percent excess = (0.3 - 0.25)/0.25 x 100 = 20%.

NaCl

= 0.5 mol x 58.5 = 29.25 g.

Practice Mole Concept More on Limiting reagent, percent yield and percent excess

Previous questionNext question

More questions on Limiting reagent, percent yield and percent excess