Question

What mass of CaCl₂ in grams would be enough to produce 14.35 gm of AgCl?

Options

• 5.55 g

• 8.29 g

• 16.59 g

• 10 g

Answer 1

5.55 g

Explanation:

\[\ce{\underset{111 g}{CaCl2} + 2AgNO3 -> Ca(NO3)2 + \underset{2 × 143.5 g}{2 AgCl}}\]

CaCl₂ require to produce 2 × 143.5 of AgCl = 111 g

CaCl₂ required to produce 14.35 g of AgCl

= `(111 xx 14.35)/(2 xx 143.5)`

= 5.55 g