Question

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what mass of cacl2 in grams would be enough to produce 14.35 gm of Agcl


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Answer 1

Question

What mass of cacl2 in grams would be enough to produce 14.35 gm of AgCl?

Solution

CaCl₂ + 2Ag → 2AgCl + 2Ca

Equivalent of CaCl₂ = Equivalent of AgCl

Eʷ = Mʷ / 2

⇒ W × 2 / 111 = 14. 35 / 143. 5

⇒ W = 5. 55 g

Answer 2

Explanation:

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