Question

What mass of calcium chloride in grams would be enough to produce 14.35 g of AgCl ? (At. mass : Ca =40 ; Ag = 108)

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

Answer 1

Answer: (A)

5.55 g

Answer 2

$\underset{\text{111 g}}{ {cacl_{2}}}+2AgNO_{3}Ca(NO_{3})_{2}+$ \underset{\text{2 \times 143.5 g}}{ {2\,Agcl}} $cacl_{2}$ required to produce $2\times143.5\,g$ of $Agcl=111\,g$ $Cacl_{2}$ required to produce $14.35\,g$ of $Agcl$ $=\frac{111\times 14.35}{2\times143.5}=5.55\,g$