MCQ
What should be the weight and moles of $AgCl$ precipitate obtained on adding $500\,ml$ of $0.20\, M$ $HCl$ in $30\, g$ of $AgN{O_3}$ solution? ($AgN{O_3}= 170$)
- ✓$14.35 \,g$
- B$15\, g$
- C$18\, g$
- D$19\, g$
✓
Answer
Correct option: A.
$14.35 \,g$
a
(a) $AgN{O_3} + HCl \to AgCl + HN{O_3}$
(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 $=14.345\,gm$
$t =t$ $0.076$ mole $0$ $0.1$ mole
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