4. \( xg \) of \( Zn \) metal was mixed with \( 2 \times g \) of iodine to form \( Znl z _{2} \). Which of the reactant will remain in excess and what fraction of it remain unreacted? (Atomic weight of \( Zn =65, l =127 \) ) (A) \( Zn , 0.75 \) (B) \( I_{2}, 0.48 \) (C) \( Zn , 0.48 \) (D) \( I_{2}, 0.75 \)

Question

4. 4. x g of Zn metal was mixed with 2x g of iodine to form Znl2. Which of the reactant will remain in excess and what fraction of it remains unreacted? (Atomic weight of Zn=65, 1=127)

(A) Zn, 0.75 

(C) Zn, 0.48 

(B) 12, 0.48

(D) 12, 0.75 ​

Answer 1

Correct option is (B) 12, 0.48

\(\because\) 1 mole of Zn react with 1 mole of I₂ to form.

1 mole ZnI₂

\(\therefore\) x/65 mole Zn required x/65 mole I₂ for complete reaction.

but we have only 2x/254 mole of I₂. it means I₂ is a limiting reagent.

\(\therefore\) Number of moles of Zn unreacted = \((\frac x{65}-\frac{2x}{254})\)

 = 0.0075x mole.

\(\therefore\) fraction of Zn unreacted = \(\frac{0.0075x}{x/65}\)

 = 0.0075 x 65 = 0.48

Comments

Thank you so much