Q. The ratio of amounts of $H_{2}S\$ needed to precipitate all the metal ions from 100 mL of 1 M ${AgNO}_{3}$ and 100mL of ${CuSO}_{4},\$ will be
A) 1 : 1
B) 1 : 2
C) 2 : 1
D) None of these
Explanation
(b)
100 mL of 1 M $AgNO_{3} \equiv 0.1$ mol $AgNO_{3}$
100 mL of 1 M $CuSO_{4} = 0.1\ mol\ CuSO_{4}$
$2AgNO_{3} + H_{2}S \rightarrow Ag_{2}S + 2HNO_{3}$
2 mol 1 mol
0.1 mol 0.05 mol
$CuSO_{4} + H_{2}S \rightarrow CuS + H_{2}SO_{4}$
1 mol 1 mol
0.1 mol 0.1 mol
$\therefore$ Ratio of the amounts of $H_{2}S$ needed=0.05:0.1=1:2
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.