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Q.
What mass of calcium chloride in grams would be enough to produce $14.35\, g$ of $AgCl$ ? (Atomic mass $Ca=40, Ag =108$)
ManipalManipal 2017
Solution:
$CaCl _{2}+2 AgNO _{3} \rightarrow Ca \left( NO _{3}\right)_{2}+2 AgCl$
$1.1\, g CaCl _{2}$ required to produce
$2 \times 143.5\, g$ of $AgCl =111\, g$
$CaCl _{2}$ required to produce $14.35\, g$ of
$AgCl =\frac{111 \times 14.35}{2 \times 143.5}$
$=5.55\, g$