What should be the weight and moles of AgCl precipitate obtained on adding 500 ml of 0.20 M HCl in 30 g of AgNO3 solution? (AgNO3 = 170)

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 AgNO3 solution? (AgNO3 = 170) (A) 14.35 g (B) 15 g (C) 18 g (D) 19 g

Tardigrade Admin · Accepted Answer

AgNO3 + HCl → AgCl + HNO3 (30/170) (500×0.2/1000) t = 0 0.176 mole 0.1 mole limiting = 14.345 g t = t 0.076 mole 0 0.1 mole

Q. What should be the weight and moles of $A g Cl$ precipitate obtained on adding $500 m l$ of $0.20 M H Cl$ in $30 g$ of $A g N O_{3}$ solution? $(A g N O_{3} = 170)$

$14.35 g$

$15 g$

$18 g$

$19 g$

$A g N O_{3} + H Cl \to A g Cl + H N O_{3}$
$\frac{30}{170}$ $\frac{500 \times 0.2}{1000}$
$t = 00.176 m o l e 0.1 m o l e l imi t in g = 14.345 g$ $t = t 0.076 m o l e 0 0.1 m o l e$

Q. What should be the weight and moles of AgCl precipitate obtained on adding 500ml of 0.20MHCl in 30g of AgNO3​ solution? (AgNO3​=170)

14.35g

15g

18g

19g