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