Question

what mass of calcium chloride in grams would be enough to produce 14.35g of Agcl

Answer 1

2Ag+ CaCl2=>AgCl+Ca
equivalents equal
$w \times \frac{2}{111} = \frac{14.35}{143.5} \\ w = 5.55$
Hope this helps you

Answer 2

2Ag+ CaCl2=>AgCl+Ca.

The value of equivalents equal and that is vital for accessing with w \times \frac{2}{111} = \frac{14.35}{143.5} \\ w = 5.55.

It is used to determine the number of moles and carry out mass by involving CaCl2.

The mass can be calculated in bringing it back by the 5.536793815 grams.