Question

What mass of calcium chloride in grams would be enough to produce $14.35$ g of AgCl?

[Atomic mass of Ca = $40$, Ag = $108$]

• A. 5.55 g
• B. 8.295 g
• C. 16.59 g
• D. 11.19 g

Answer 1

The correct option is A 5.55 g

$CaC{l}_2+2Ag\to 2AgCl+2Ca$

Equivalent of $CaC{l}_2$ = Equivalent of $AgCl$

$\therefore \frac{w\times 2}{111}=\frac{14.35}{143.5}[E_w=\frac{Mw}{2}]$

$w=5.55g$.