A) 14.35 g
B) 15 g
C) 18 g
D) 19 g
Correct Answer: A
Solution :
[a] \[AgN{{O}_{3}}+HCl\to AgCl+HN{{O}_{3}}\] |
\[\frac{30}{170}\] \[\frac{500\times 0.2}{1000}\] |
t =0 0.176 mole 0.1 mole limiting =14.345gm |
t =t 0.076 mole 0 0.1mole |
You need to login to perform this action.
You will be redirected in
3 sec