what mass of calcium chloride in grams would be enough to produce 14.35g of silver chloride
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CaCl₂ + 2Ag → 2AgCl + 2Ca
✓ Equivalent of CaCl₂ = Equivalent of AgCl
Eʷ = Mʷ / 2
⇒ w × 2 / 111 = 14. 35 / 143. 5
⇒ W = 5. 55 g
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CaCl₂ + 2Ag → 2AgCl + 2Ca
✓ Equivalent of CaCl₂ = Equivalent of AgCl
Eʷ = Mʷ / 2
⇒ w × 2 / 111 = 14. 35 / 143. 5
⇒ W = 5. 55 g
Cheers!
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