Question

What should be the weight and moles of $AgCl$ precipitate obtained on adding $500 \,ml$ of $0.20 \,M \,HCl$ in $30 \,g$ of $AgNO_3$ solution? $\left(AgNO_3 = 170\right)$

• A. $14.35\, g$
• B. $15\, g$
• C. $18\,g$
• D. $19\, g$

Answer 1

Answer: (A)

$AgNO_3 + HCl \to AgCl + HNO_3$
$\frac{30}{170}$ $\frac{500\times0.2}{1000}$
$t = 0 0.176 \,mole\, 0.1\, mole \,limiting \,=\,14.345\, g$ $t\, =\, t\, 0.076\, mole\, 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.1\,mole$