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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» 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`


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