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*19g`

Answer 1

Correct Answer - A
`underset(111g)(CaCl_(2)) + 2AgNO_(3) to Ca(NO_(3))_(2)+underset(2xx143.5 g)(2AgCl)`
`CaCl_(2)` required to produced `2xx143*5g` of AgCl `= 111g`
`CaCl_(2)` required to produce `14*35g` of AgCl
`=(111xx14.35)/(2xx143.5)=5*55g`